4x+2x^2=396

Solution for 4x+2x^2=396 equation:

4x+2x^2=396

We move all terms to the left:

4x+2x^2-(396)=0

a = 2; b = 4; c = -396;
Δ = b2-4ac
Δ = 42-4·2·(-396)
Δ = 3184
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:

$\sqrt{\Delta}=\sqrt{3184}=\sqrt{16*199}=\sqrt{16}*\sqrt{199}=4\sqrt{199}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{199}}{2*2}=\frac{-4-4\sqrt{199}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{199}}{2*2}=\frac{-4+4\sqrt{199}}{4}
$