2x+2x^2=216

Solution for 2x+2x^2=216 equation:

2x+2x^2=216

We move all terms to the left:

2x+2x^2-(216)=0

a = 2; b = 2; c = -216;
Δ = b2-4ac
Δ = 22-4·2·(-216)
Δ = 1732
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:

$\sqrt{\Delta}=\sqrt{1732}=\sqrt{4*433}=\sqrt{4}*\sqrt{433}=2\sqrt{433}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{433}}{2*2}=\frac{-2-2\sqrt{433}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{433}}{2*2}=\frac{-2+2\sqrt{433}}{4}
$