a^2+8a-24=0

Solution for a^2+8a-24=0 equation:

a^2+8a-24=0

a = 1; b = 8; c = -24;
Δ = b2-4ac
Δ = 82-4·1·(-24)
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{10}}{2*1}=\frac{-8-4\sqrt{10}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{10}}{2*1}=\frac{-8+4\sqrt{10}}{2}
$