3r^2+5070r-240300=0

Solution for 3r^2+5070r-240300=0 equation:

3r^2+5070r-240300=0

a = 3; b = 5070; c = -240300;
Δ = b2-4ac
Δ = 50702-4·3·(-240300)
Δ = 28588500
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:

$\sqrt{\Delta}=\sqrt{28588500}=\sqrt{900*31765}=\sqrt{900}*\sqrt{31765}=30\sqrt{31765}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5070)-30\sqrt{31765}}{2*3}=\frac{-5070-30\sqrt{31765}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5070)+30\sqrt{31765}}{2*3}=\frac{-5070+30\sqrt{31765}}{6}
$