aryabhatt's quadratic formula
5r2=80
you need it in carpentry
In problems of motion, especially involving constant acceleration, a quadratic equation will from the formulas of motion to solve for time, usually. This is just one example.
The quadratic formula always works (as long as one considers complex numbers). "Simple rearrangement" may be quicker when the numbers look simple enough for you to decide (or rather guess) what the factors/ roots are by inspection (but the "rearrangement" method still works -- the numbers may just be more complicated). Probably the easiest quadratic is when the coefficient of x is zero (i.e. a polynomial of the form ax^2+b=0) or when there is no constant term (i.e. ax^2+bx=0) The quadratic formula cannot be used to solve an equation if a term in the equation has a degree higher than 2 (or if it can't be put in the form ax^2+bx+c=0). There are other more complex formulas for polynomials for degree 3 and 4.
excel formulas multipy and divide first if i remember corectly
5r2=80
quadratic formula
ax2 +bx + c To find roots of any quadratic equation. X = - b (+/-) sqrt(b2 - 4ac)/2a
you need it in carpentry
In problems of motion, especially involving constant acceleration, a quadratic equation will from the formulas of motion to solve for time, usually. This is just one example.
is this what you were looking for? there are many different types of quadratic formulas-- -b √ b^2 - 4ac = x (over) 2a
at first the first person to solve the quadratic equation is from the middle kingdom of Egypt. Greeks were also able to solve the quadratic equation but that was on the unproper way. Greeks were able to solve the quadratic equation by geometric method or equlid's method. equlid's method contains only three quadratic equation. dipohantus have also solved the quadratic equations but he have solved by giving only two roots any they both were only of positive signs.After that arbhatya also gave the two formulas for quadratic equation but the bentaguptahave only accepted only one of them after theat some of the Indian mathematican have also solved the quadratic equation who gave the proper definations and formula and in this way quadratic equation have been formed. Prabesh Regmi Kanjirowa National School
Probably not. Accounting doesn't use much math beyond basic additions, subtractions, multiplications, and divisions.
To solve a binomial expression, you typically simplify or factor it. If you're solving an equation set to zero, you can use methods like factoring, completing the square, or applying the quadratic formula if it's a quadratic binomial. For binomials, you may also apply the difference of squares or the sum/difference of cubes formulas if applicable. Always ensure to check your solutions by substituting them back into the original expression.
The quadratic formula always works (as long as one considers complex numbers). "Simple rearrangement" may be quicker when the numbers look simple enough for you to decide (or rather guess) what the factors/ roots are by inspection (but the "rearrangement" method still works -- the numbers may just be more complicated). Probably the easiest quadratic is when the coefficient of x is zero (i.e. a polynomial of the form ax^2+b=0) or when there is no constant term (i.e. ax^2+bx=0) The quadratic formula cannot be used to solve an equation if a term in the equation has a degree higher than 2 (or if it can't be put in the form ax^2+bx+c=0). There are other more complex formulas for polynomials for degree 3 and 4.
Parabolas are used in real life in light reflectors on cars to create a concentrated beam of intense light. Braking distance and stopping distance are quadratic formulas so their graphs are parabolas. A ball in motion in space has a path of a parabola.
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