x+4x+x^2=200

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

x+4x+x^2=200

We move all terms to the left:

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

We add all the numbers together, and all the variables

x^2+5x-200=0

a = 1; b = 5; c = -200;
Δ = b2-4ac
Δ = 52-4·1·(-200)
Δ = 825
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:

$\sqrt{\Delta}=\sqrt{825}=\sqrt{25*33}=\sqrt{25}*\sqrt{33}=5\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{33}}{2*1}=\frac{-5-5\sqrt{33}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{33}}{2*1}=\frac{-5+5\sqrt{33}}{2}
$