x(2x-12)=40

Solution for x(2x-12)=40 equation:

x(2x-12)=40

We move all terms to the left:

x(2x-12)-(40)=0

We multiply parentheses

2x^2-12x-40=0

a = 2; b = -12; c = -40;
Δ = b2-4ac
Δ = -122-4·2·(-40)
Δ = 464
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:

$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{29}}{2*2}=\frac{12-4\sqrt{29}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{29}}{2*2}=\frac{12+4\sqrt{29}}{4}
$