(2x^2)+(2x)+4=1252

Solution for (2x^2)+(2x)+4=1252 equation:

(2x^2)+(2x)+4=1252

We move all terms to the left:

(2x^2)+(2x)+4-(1252)=0

We add all the numbers together, and all the variables

2x^2+2x-1248=0

a = 2; b = 2; c = -1248;
Δ = b2-4ac
Δ = 22-4·2·(-1248)
Δ = 9988
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:

$\sqrt{\Delta}=\sqrt{9988}=\sqrt{4*2497}=\sqrt{4}*\sqrt{2497}=2\sqrt{2497}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{2497}}{2*2}=\frac{-2-2\sqrt{2497}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{2497}}{2*2}=\frac{-2+2\sqrt{2497}}{4}
$