r^2+16r+10=0

Solution for r^2+16r+10=0 equation:

r^2+16r+10=0

a = 1; b = 16; c = +10;
Δ = b2-4ac
Δ = 162-4·1·10
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-6\sqrt{6}}{2*1}=\frac{-16-6\sqrt{6}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+6\sqrt{6}}{2*1}=\frac{-16+6\sqrt{6}}{2}
$