2x+4x^2=400

Solution for 2x+4x^2=400 equation:

2x+4x^2=400

We move all terms to the left:

2x+4x^2-(400)=0

a = 4; b = 2; c = -400;
Δ = b2-4ac
Δ = 22-4·4·(-400)
Δ = 6404
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:

$\sqrt{\Delta}=\sqrt{6404}=\sqrt{4*1601}=\sqrt{4}*\sqrt{1601}=2\sqrt{1601}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{1601}}{2*4}=\frac{-2-2\sqrt{1601}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{1601}}{2*4}=\frac{-2+2\sqrt{1601}}{8}
$