x^2-4x=5x+108

Solution for x^2-4x=5x+108 equation:

x^2-4x=5x+108

We move all terms to the left:

x^2-4x-(5x+108)=0

We get rid of parentheses

x^2-4x-5x-108=0

We add all the numbers together, and all the variables

x^2-9x-108=0

a = 1; b = -9; c = -108;
Δ = b2-4ac
Δ = -92-4·1·(-108)
Δ = 513
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:

$\sqrt{\Delta}=\sqrt{513}=\sqrt{9*57}=\sqrt{9}*\sqrt{57}=3\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3\sqrt{57}}{2*1}=\frac{9-3\sqrt{57}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3\sqrt{57}}{2*1}=\frac{9+3\sqrt{57}}{2}
$