2x^2+2x=440

Solution for 2x^2+2x=440 equation:

2x^2+2x=440

We move all terms to the left:

2x^2+2x-(440)=0

a = 2; b = 2; c = -440;
Δ = b2-4ac
Δ = 22-4·2·(-440)
Δ = 3524
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:

$\sqrt{\Delta}=\sqrt{3524}=\sqrt{4*881}=\sqrt{4}*\sqrt{881}=2\sqrt{881}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{881}}{2*2}=\frac{-2-2\sqrt{881}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{881}}{2*2}=\frac{-2+2\sqrt{881}}{4}
$