252=28x+2x^2

Solution for 252=28x+2x^2 equation:

252=28x+2x^2

We move all terms to the left:

252-(28x+2x^2)=0

We get rid of parentheses

-2x^2-28x+252=0

a = -2; b = -28; c = +252;
Δ = b2-4ac
Δ = -282-4·(-2)·252
Δ = 2800
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:

$\sqrt{\Delta}=\sqrt{2800}=\sqrt{400*7}=\sqrt{400}*\sqrt{7}=20\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-20\sqrt{7}}{2*-2}=\frac{28-20\sqrt{7}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+20\sqrt{7}}{2*-2}=\frac{28+20\sqrt{7}}{-4}
$