2x^2+10x=936

Solution for 2x^2+10x=936 equation:

2x^2+10x=936

We move all terms to the left:

2x^2+10x-(936)=0

a = 2; b = 10; c = -936;
Δ = b2-4ac
Δ = 102-4·2·(-936)
Δ = 7588
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:

$\sqrt{\Delta}=\sqrt{7588}=\sqrt{4*1897}=\sqrt{4}*\sqrt{1897}=2\sqrt{1897}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{1897}}{2*2}=\frac{-10-2\sqrt{1897}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{1897}}{2*2}=\frac{-10+2\sqrt{1897}}{4}
$