r^2+16r-160=0

Solution for r^2+16r-160=0 equation:

r^2+16r-160=0

a = 1; b = 16; c = -160;
Δ = b2-4ac
Δ = 162-4·1·(-160)
Δ = 896
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{896}=\sqrt{64*14}=\sqrt{64}*\sqrt{14}=8\sqrt{14}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{14}}{2*1}=\frac{-16-8\sqrt{14}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{14}}{2*1}=\frac{-16+8\sqrt{14}}{2}
$