3x(x-12)=40

Solution for 3x(x-12)=40 equation:

3x(x-12)=40

We move all terms to the left:

3x(x-12)-(40)=0

We multiply parentheses

3x^2-36x-40=0

a = 3; b = -36; c = -40;
Δ = b2-4ac
Δ = -362-4·3·(-40)
Δ = 1776
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1776}=\sqrt{16*111}=\sqrt{16}*\sqrt{111}=4\sqrt{111}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4\sqrt{111}}{2*3}=\frac{36-4\sqrt{111}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4\sqrt{111}}{2*3}=\frac{36+4\sqrt{111}}{6}
$