How do you solve $x^2+3x=12$ using the quadratic formula?

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Answer 1 · 2017-05-15T17:58:57

$color(blue)(x=2.275$ or $color(blue)(x=-5.275$ to the nearest 3 decimal places

quadratic formula:
$x=(-b+-sqrt(b^2-4ac))/(2a)$

Quadratic equation: $ax^2 +bx+c =0$

$x^2+3x=12$

$:.x^2+3x-12=0$

$:.a=1,b=3,c=-12$

$:.x=(-(3)+-sqrt((3)^2-4(1)(-12)))/(2(1))$

$:.x=(-3+-sqrt(9+48))/(2(1))$

$:.x=(-3+-sqrt(57))/(2)$

$:.x=(-3+7.549834435)/2$ or $x=(-3-7.549834435)/2$

$:.x=(4.549834435)/2$ or $x=(-10.549834435)/2$

$:.x=2.2749172175$ or $x=-5.2749172175$

$:.color(blue)(x=2.275$ or $color(blue)(x=-5.275$ to the nearest 3 decimal places

substitute $color(blue)(x=2.2749172175$

$:.(2.2749172175)^2+3(2.2749172175)=12$

$:.5.175248344+6.824751651=12$

$:.12=12$

substitute $color(blue)(x=-5.2749172175$

$:.(-5.2749172175)^2+3(-5.2749172175)=12$

$:.27.82475165-15.82475165=12$

$:.12=12$

How do you solve x^2+3x=12x^2+3x=12 using the quadratic formula?