5r^2+37r-24=0

Solution for 5r^2+37r-24=0 equation:

5r^2+37r-24=0

a = 5; b = 37; c = -24;
Δ = b2-4ac
Δ = 372-4·5·(-24)
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(37)-43}{2*5}=\frac{-80}{10}
=-8
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(37)+43}{2*5}=\frac{6}{10}
=3/5
$