16x+x^2=180

Solution for 16x+x^2=180 equation:

16x+x^2=180

We move all terms to the left:

16x+x^2-(180)=0

a = 1; b = 16; c = -180;
Δ = b2-4ac
Δ = 162-4·1·(-180)
Δ = 976
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:

$\sqrt{\Delta}=\sqrt{976}=\sqrt{16*61}=\sqrt{16}*\sqrt{61}=4\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{61}}{2*1}=\frac{-16-4\sqrt{61}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{61}}{2*1}=\frac{-16+4\sqrt{61}}{2}
$