How do you solve $4x^2-13x=12$ ?

Question

2978 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-27T15:44:40

First subtract $12$ from both sides to get $4x^2-13x-12 = 0$ , then use the quadratic formula to find:

$x = (13 +- 19)/8$

That is $x=4$ or $x=-3/4$

$4x^2-13x-12$ is in the form $ax^2+bx+c$ with $a=4$ , $b=-13$ and $c=-12$

It discriminant is given by the formula:

$Delta = b^2-4ac = 13^2 - (4xx4xx-12) = 169+192 = 361 = 19^2$

Since $Delta$ is positive and a perfect square, the roots of our quadratic are rational, given by the formula:

$x = (-b+-sqrt(Delta))/(2a) = (13+-19)/8$

That is $x=32/8=4$ or $x=-6/8 = -3/4$

How do you solve 4x^2-13x=124x^2-13x=12?