r^2+10r-16=0

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

r^2+10r-16=0

a = 1; b = 10; c = -16;
Δ = b2-4ac
Δ = 102-4·1·(-16)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{41}}{2*1}=\frac{-10-2\sqrt{41}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{41}}{2*1}=\frac{-10+2\sqrt{41}}{2}
$