4x^2+4x=198

Solution for 4x^2+4x=198 equation:

4x^2+4x=198

We move all terms to the left:

4x^2+4x-(198)=0

a = 4; b = 4; c = -198;
Δ = b2-4ac
Δ = 42-4·4·(-198)
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*4}=\frac{-4-4\sqrt{199}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{199}}{2*4}=\frac{-4+4\sqrt{199}}{8}
$