3v^2+150v-2500=0

Solution for 3v^2+150v-2500=0 equation:

3v^2+150v-2500=0

a = 3; b = 150; c = -2500;
Δ = b2-4ac
Δ = 1502-4·3·(-2500)
Δ = 52500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{52500}=\sqrt{2500*21}=\sqrt{2500}*\sqrt{21}=50\sqrt{21}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(150)-50\sqrt{21}}{2*3}=\frac{-150-50\sqrt{21}}{6}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(150)+50\sqrt{21}}{2*3}=\frac{-150+50\sqrt{21}}{6}
$