828=(2x)^2+10x

Solution for 828=(2x)^2+10x equation:

828=(2x)^2+10x

We move all terms to the left:

828-((2x)^2+10x)=0

We get rid of parentheses

-2x^2-10x+828=0

a = -2; b = -10; c = +828;
Δ = b2-4ac
Δ = -102-4·(-2)·828
Δ = 6724
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6724}=82$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-82}{2*-2}=\frac{-72}{-4}
=+18
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+82}{2*-2}=\frac{92}{-4}
=-23
$