4x+2x^2=180

Solution for 4x+2x^2=180 equation:

4x+2x^2=180

We move all terms to the left:

4x+2x^2-(180)=0

a = 2; b = 4; c = -180;
Δ = b2-4ac
Δ = 42-4·2·(-180)
Δ = 1456
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:

$\sqrt{\Delta}=\sqrt{1456}=\sqrt{16*91}=\sqrt{16}*\sqrt{91}=4\sqrt{91}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{91}}{2*2}=\frac{-4-4\sqrt{91}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{91}}{2*2}=\frac{-4+4\sqrt{91}}{4}
$