How to solve a quadratic equation? step by step
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2 Answers

yuze quadratik equashun

start: ax^2+bx+c=0

first get sqrt(b^2-4ac)...math-man kall that the "diskriminant"

roots=-b/2a +-diskriminant/2a

zampel...x^2+x-72=0

a=1, b=1, c=-72

b^2-4ac=289, sqrt(289)=17,  17/2a=8.5

-b/2a=-0.5

anser=0.5+-8.5....8 & -9
by
There are two ways to do this. You can apply the formula or complete the square (if you've forgotten the formula). Completing the square gives us the formula anyway. We start with a typical quadratic written ax^2 + bx + c = 0. Divide the equation by a so that we simplify the x^2, we get x^2 + (b/a)x + (c/a) = 0. What we're trying to do now is to get this to look like (a perfect square) + constant = 0, because that's easy to solve, assuming there is a solution. Now the square of x + A = x^2 + 2Ax + A^2. So, if 2A = (b/a), A = (b/2a) and A^2 = (b^2/4a^2). This transforms our equation into (x + (b/2a))^2  - (b^2/4a^2) + c/a = 0. We have to subtract the A^2 term from the constant because we added it to complete the square. So (x + (b/2a))^2 = (b^2/4a^2) - c/a = (b^2 - 4ac)/4a^2. Taking square roots of both sides we get x + (b/2a) = sqrt(b^2 - 4ac) /2a. Sqrt can be positive or negative +/- so x = +/-(b^2 - 4ac) /2a - (b/2a). This is usually written (-b +/- sqrt(b^2 - 4ac))/2a.
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