2x+x2=169

Solution for 2x+x2=169 equation:

2x+x2=169

We move all terms to the left:

2x+x2-(169)=0

We add all the numbers together, and all the variables

x^2+2x-169=0

a = 1; b = 2; c = -169;
Δ = b2-4ac
Δ = 22-4·1·(-169)
Δ = 680
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:

$\sqrt{\Delta}=\sqrt{680}=\sqrt{4*170}=\sqrt{4}*\sqrt{170}=2\sqrt{170}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{170}}{2*1}=\frac{-2-2\sqrt{170}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{170}}{2*1}=\frac{-2+2\sqrt{170}}{2}
$