15r^2+150r-300=0

Solution for 15r^2+150r-300=0 equation:

15r^2+150r-300=0

a = 15; b = 150; c = -300;
Δ = b2-4ac
Δ = 1502-4·15·(-300)
Δ = 40500
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:

$\sqrt{\Delta}=\sqrt{40500}=\sqrt{8100*5}=\sqrt{8100}*\sqrt{5}=90\sqrt{5}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(150)-90\sqrt{5}}{2*15}=\frac{-150-90\sqrt{5}}{30}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(150)+90\sqrt{5}}{2*15}=\frac{-150+90\sqrt{5}}{30}
$