38x+x^2=180

Solution for 38x+x^2=180 equation:

38x+x^2=180

We move all terms to the left:

38x+x^2-(180)=0

a = 1; b = 38; c = -180;
Δ = b2-4ac
Δ = 382-4·1·(-180)
Δ = 2164
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:

$\sqrt{\Delta}=\sqrt{2164}=\sqrt{4*541}=\sqrt{4}*\sqrt{541}=2\sqrt{541}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-2\sqrt{541}}{2*1}=\frac{-38-2\sqrt{541}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+2\sqrt{541}}{2*1}=\frac{-38+2\sqrt{541}}{2}
$