3v^2+15v-48=0

Solution for 3v^2+15v-48=0 equation:

3v^2+15v-48=0

a = 3; b = 15; c = -48;
Δ = b2-4ac
Δ = 152-4·3·(-48)
Δ = 801
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:

$\sqrt{\Delta}=\sqrt{801}=\sqrt{9*89}=\sqrt{9}*\sqrt{89}=3\sqrt{89}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3\sqrt{89}}{2*3}=\frac{-15-3\sqrt{89}}{6}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3\sqrt{89}}{2*3}=\frac{-15+3\sqrt{89}}{6}
$