The roots of the equation are -4 and 3. Simplest solution is (x +4 )(x - 3) which multiplies out as x^2 + x - 12 = 0
Yes because rearranging it into the form of 3x2-10x+5 = 0 makes the discriminant of the quadratic equation greater than zero which means it will have two different solutions. Solving the equation by means of the quadratic formula gives x as being 2.721 or 0.613 both corrected to 3 decimal places.
That's not an equation - it doesn't have an equal sign. Assuming you mean 2x2 - 3x - 90 = 0, you can find the solution, or usually the two solutions, of such equations with the quadratic formula. In this case, replace a = 2, b = -3, c = -90.
To find the zeros of this quadratic function, y= 3x^2 + 6x - 9, we must equal y to 0. So we have the quadratic equation: 3x^2+6x-9 = 0, where a = 3, b = 6, and c = -9 The quadratic formula: x = [-b ± √(b^2 - 4ac)]/(2a) substitute what you know into this formula; x = [-6 ± √(6^2 - 4 x 3 x -9)]/(2 x 3) x = [-6 ± √(36 +108)]/6 x = (-6 ± √144)/6 x = (-6 ± 12)/6 Simplify: mulyiply by 1/6 both the numerator and the denominator; x = -1 ± 2 x = -1 + 2 or x = -1 - 2 x = 1 or x = -3 So solutions are -3 and 1. If you check the answers by plugging them into the equation, you will see that they work.
Without an equality sign and not knowing the plus or minus values of 3x and 3 the information given can't be considered to be a quadratic equation.
There is a new method, called Diagonal Sum Method, that quickly and directly give the 2 roots without having to factor the equation. The innovative concept of this method is finding 2 fractions knowing their sum (-b/a) and their product (c/a). It is fast, convenient and is applicable to any quadratic equation in standard form ax^2 + bx + c = 0, whenever it can be factored. If it fails to find answer, then the equation is not factorable, and consequently, the quadratic formula must be used. So, I advise you to proceed solving any quadratic equation in 2 steps. First, find out if the equation can be factored? How?. Use this new method to solve it. It usually takes fewer than 3 trials. If its fails then use the quadratic formula to solve it in the second step. See book titled:" New methods for solving quadratic equations and inequalities" (Trafford Publishing 2009)
The quadratic formula can be used to solve an equation only if the highest degree in the equation is 2.
y=2(x-3)+1
I suggest you use the quadratic formula. In this case, a = 1, b = 5, c = 3.
This does not factor to whole numbers. You need to use the quadratic formula... -b+-sqrt[b2-4(a)(c)] / 2a Your equation is this [below]. Identify what a,b and c equals and put it into your quadratic formula. 3m2-88 a=3 b=0 c=-88 when you work this out you get two answers but they are not whole numbers. your answers are 5.416 an -5.416
It is a quadratic equation that can be solved by using the quadratic equation formula whereas x = -9.321825 or x = 0.321825 both given to 6 decimal places
By using the quadratic equation formula: (4x+1)(x-3) = 0
If you mean: 11x2-34x+3 = 0 then the solutions are x = 1/11 and x = 3 by completing the square or using the quadratic equation formula
Using the quadratic equation formula: x = -3 - the square root of 3 or x = -3 + the square root of 3
Use the quadratic formula, with a = 1, b = -3, c = 2.
56w2-17w-3 = (8w+1)(7w-3) Using the quadratic equation formula helps.
To solve the quadratic equation, S^2 + 4S - 21 = 0, you can factor the expression or use the quadratic formula. Factoring, we can rewrite it as (S-3)(S+7) = 0. This means that either S-3 = 0 or S+7 = 0. Solving for S in each case gives S = 3 or S = -7 as the solutions to the equation.
Roots, zeroes, and x values are 3 other names for solutions of a quadratic equation.