4v^2=6v+15

Solution for 4v^2=6v+15 equation:

4v^2=6v+15

We move all terms to the left:

4v^2-(6v+15)=0

We get rid of parentheses

4v^2-6v-15=0

a = 4; b = -6; c = -15;
Δ = b2-4ac
Δ = -62-4·4·(-15)
Δ = 276
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:

$\sqrt{\Delta}=\sqrt{276}=\sqrt{4*69}=\sqrt{4}*\sqrt{69}=2\sqrt{69}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{69}}{2*4}=\frac{6-2\sqrt{69}}{8}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{69}}{2*4}=\frac{6+2\sqrt{69}}{8}
$