5r^2+8r-24=0

Solution for 5r^2+8r-24=0 equation:

5r^2+8r-24=0

a = 5; b = 8; c = -24;
Δ = b2-4ac
Δ = 82-4·5·(-24)
Δ = 544
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:

$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{34}}{2*5}=\frac{-8-4\sqrt{34}}{10}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{34}}{2*5}=\frac{-8+4\sqrt{34}}{10}
$