2x^2+10x=108

Solution for 2x^2+10x=108 equation:

2x^2+10x=108

We move all terms to the left:

2x^2+10x-(108)=0

a = 2; b = 10; c = -108;
Δ = b2-4ac
Δ = 102-4·2·(-108)
Δ = 964
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:

$\sqrt{\Delta}=\sqrt{964}=\sqrt{4*241}=\sqrt{4}*\sqrt{241}=2\sqrt{241}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{241}}{2*2}=\frac{-10-2\sqrt{241}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{241}}{2*2}=\frac{-10+2\sqrt{241}}{4}
$