r^2+10r-96=0

Solution for r^2+10r-96=0 equation:

r^2+10r-96=0

a = 1; b = 10; c = -96;
Δ = b2-4ac
Δ = 102-4·1·(-96)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-22}{2*1}=\frac{-32}{2}
=-16
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+22}{2*1}=\frac{12}{2}
=6
$