How do you solve $x^2+x-12=0$ using the quadratic formula?

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Answer 1 · 2018-05-02T19:23:10

The answer $x_1=3$

$x_2=-4$

$ax^2+bx+c=0$

$x^2+x-12=0$

$a=1,b=1andc=-12$

$x=[-b+-sqrt(b^2-4ac)]/(2a)$

$x=[-1+-sqrt(1+48)]/(2)$

$x=[-1+-7]/(2)$

$x_1=(-1+7)/2=6/2=3$

$x_2=(-1-7)/2=-8/2=-4$

How do you solve x^2+x-12=0x^2+x-12=0 using the quadratic formula?