4x^2+20x+16=200

Solution for 4x^2+20x+16=200 equation:

4x^2+20x+16=200

We move all terms to the left:

4x^2+20x+16-(200)=0

We add all the numbers together, and all the variables

4x^2+20x-184=0

a = 4; b = 20; c = -184;
Δ = b2-4ac
Δ = 202-4·4·(-184)
Δ = 3344
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:

$\sqrt{\Delta}=\sqrt{3344}=\sqrt{16*209}=\sqrt{16}*\sqrt{209}=4\sqrt{209}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{209}}{2*4}=\frac{-20-4\sqrt{209}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{209}}{2*4}=\frac{-20+4\sqrt{209}}{8}
$