(3x^2-11x+12)-(18x^2+20x-100)=0

Solution for (3x^2-11x+12)-(18x^2+20x-100)=0 equation:

(3x^2-11x+12)-(18x^2+20x-100)=0

We get rid of parentheses

3x^2-18x^2-11x-20x+12+100=0

We add all the numbers together, and all the variables

-15x^2-31x+112=0

a = -15; b = -31; c = +112;
Δ = b2-4ac
Δ = -312-4·(-15)·112
Δ = 7681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-\sqrt{7681}}{2*-15}=\frac{31-\sqrt{7681}}{-30}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{7681}}{2*-15}=\frac{31+\sqrt{7681}}{-30}
$