x^2+16x=161

Solution for x^2+16x=161 equation:

x^2+16x=161

We move all terms to the left:

x^2+16x-(161)=0

a = 1; b = 16; c = -161;
Δ = b2-4ac
Δ = 162-4·1·(-161)
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-30}{2*1}=\frac{-46}{2}
=-23
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+30}{2*1}=\frac{14}{2}
=7
$