Best Answer

if you have 3*x^2 + 2 * x + 3 = 4, you can rewrite it to

3*x^2 + 2 * x - 1 = 0

edit:

example:

2x**2-15x-15=15+2x+x**2

subtract away from one side

2x**2-15x-15-x**2=15-2x+x**2-x**2

x**2-15x-15=15-2x

x**2-15x-15-15=15-2x-15

x**2-15x-30=-2x

x**2-15x-30+2x=-2x+2x

x**2-13x-30=0

Factor (long way):

-mulitply first and third terms:

1*-30=-30

-factors of -30 that add up to second term:

30=1*2*3*5

-30=-1*30, 1*-30

-30=-2*15, 2*-15

-30=-3*10, 3*-10

-30=-5*6, 5*-6

-rewrite polynomial:

x**2-13x-30=0

x**2-15x+2x-30=0

-factor sets of terms (ie first and second, third and fourth):

x**2-15x+2x-30=0

x(x-15)+2(x-15)=0

-factor:

x(x-15)+2(x-15)=0

(x+2)(x-15)=0

-determine roots:

x+2=0

x+2-2=0-2

x=-2

x-15=0

x-15+15=0+15

x=15

Factor (short way):

x**2-13x-30=0

-because the first term is x**2 you can:

1. Take third term and factor, choose factors that add up to second term:

30=1*2*3*5

-30=-1*30, 1*-30

-30=-2*15, 2*-15

-30=-3*10, 3*-10

-30=-5*6, 5*-6

2. Place factors in (x+a)(x+b) where a and b are factors of the third term whose sum is the second term

(x+2)(x-15)

3. Determine roots:

x+2=0

x+2-2=0-2

x=-2

x-15=0

x-15+15=0+15

x=15

Q: How do you find the roots of quadratic equations that are not ending in zero?

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Well, that depends on what you mean "solve by factoring." For any quadratic equation, it is possible to factor the quadratic, and then the roots can be recovered from the factors. So in the very weak sense that every quadratic can be solved by a method that involves getting the factors and recovering the roots from them, all quadratic equations can be solved by factoring. However, in most cases, the only way of factoring the quadratic in the first place is to first find out what its roots are, and then use the roots to factor the quadratic (any quadratic polynomial can be factored as k(x - r)(x - s), where k is the leading coefficient of the polynomial and r and s are its two roots), in which case trying to recover the roots from the factors is redundant (since you had to know what the roots were to get the factors in the first place). So to really count as solving by factoring, it makes sense to require that the solution method obtains the factors by means that _don't_ require already knowing the roots of the polynomial. And in this sense, most quadratic equations are not solvable through factoring.

To find the roots (solutions) of a quadratic equation.

You substitute the value of the variable into the quadratic equation and evaluate the expression.

ve can find out quadratic equations

There are an infinite number of different quadratic equations. The quadratic formula is a single formula that can be used to find the pair of solutions to every quadratic equation.

Related questions

In general, quadratic equations have graphs that are parabolas. The quadratic formula tells us how to find the roots of a quadratic equations. If those roots are real, they are the x intercepts of the parabola.

ax+b=)

Well, that depends on what you mean "solve by factoring." For any quadratic equation, it is possible to factor the quadratic, and then the roots can be recovered from the factors. So in the very weak sense that every quadratic can be solved by a method that involves getting the factors and recovering the roots from them, all quadratic equations can be solved by factoring. However, in most cases, the only way of factoring the quadratic in the first place is to first find out what its roots are, and then use the roots to factor the quadratic (any quadratic polynomial can be factored as k(x - r)(x - s), where k is the leading coefficient of the polynomial and r and s are its two roots), in which case trying to recover the roots from the factors is redundant (since you had to know what the roots were to get the factors in the first place). So to really count as solving by factoring, it makes sense to require that the solution method obtains the factors by means that _don't_ require already knowing the roots of the polynomial. And in this sense, most quadratic equations are not solvable through factoring.

To find the roots (solutions) of a quadratic equation.

If the quadratic is ax2 + bx + c = 0 then the product of the roots is c/a.

You substitute the value of the variable into the quadratic equation and evaluate the expression.

Write an algorithm to find the root of quadratic equation

ve can find out quadratic equations

The quadratic formula is used today to find the solutions to quadratic equations, which are equations of the form ax^2 + bx + c = 0. By using the quadratic formula, we can determine the values of x that satisfy the quadratic equation and represent the points where the graph of the equation intersects the x-axis.

There are an infinite number of different quadratic equations. The quadratic formula is a single formula that can be used to find the pair of solutions to every quadratic equation.

When you need to find the roots of a quadratic equation and factorisation does not work (or you cannot find the factors). The quadratic equation ALWAYS works. And when appropriate, it will give the imaginary roots which, judging by this question, you may not yet be ready for.

The quadratic formula is used all the time to solve quadratic equations, often when the factors are fractions or decimals but sometimes as the first choice of solving method. The quadratic formula is sometimes faster than completing the square or any other factoring methods. Quadratic formula find: -x-intercept -where the parabola cross the x-axis -roots -solutions