x^2+2x^+x=198

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

x^2+2x^+x=198

We move all terms to the left:

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

We add all the numbers together, and all the variables

x^2+3x-198=0

a = 1; b = 3; c = -198;
Δ = b2-4ac
Δ = 32-4·1·(-198)
Δ = 801
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:

$\sqrt{\Delta}=\sqrt{801}=\sqrt{9*89}=\sqrt{9}*\sqrt{89}=3\sqrt{89}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{89}}{2*1}=\frac{-3-3\sqrt{89}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{89}}{2*1}=\frac{-3+3\sqrt{89}}{2}
$