r^2+32r+5=0

Solution for r^2+32r+5=0 equation:

r^2+32r+5=0

a = 1; b = 32; c = +5;
Δ = b2-4ac
Δ = 322-4·1·5
Δ = 1004
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:

$\sqrt{\Delta}=\sqrt{1004}=\sqrt{4*251}=\sqrt{4}*\sqrt{251}=2\sqrt{251}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-2\sqrt{251}}{2*1}=\frac{-32-2\sqrt{251}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+2\sqrt{251}}{2*1}=\frac{-32+2\sqrt{251}}{2}
$