3x^2+8x=x^2-3x-12

Solution for 3x^2+8x=x^2-3x-12 equation:

3x^2+8x=x^2-3x-12

We move all terms to the left:

3x^2+8x-(x^2-3x-12)=0

We get rid of parentheses

3x^2-x^2+8x+3x+12=0

We add all the numbers together, and all the variables

2x^2+11x+12=0

a = 2; b = 11; c = +12;
Δ = b2-4ac
Δ = 112-4·2·12
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-5}{2*2}=\frac{-16}{4}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+5}{2*2}=\frac{-6}{4}
=-1+1/2
$