r^2+16r-40=0

Solution for r^2+16r-40=0 equation:

r^2+16r-40=0

a = 1; b = 16; c = -40;
Δ = b2-4ac
Δ = 162-4·1·(-40)
Δ = 416
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:

$\sqrt{\Delta}=\sqrt{416}=\sqrt{16*26}=\sqrt{16}*\sqrt{26}=4\sqrt{26}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{26}}{2*1}=\frac{-16-4\sqrt{26}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{26}}{2*1}=\frac{-16+4\sqrt{26}}{2}
$