4v^2=15v+4

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

4v^2=15v+4

We move all terms to the left:

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

We get rid of parentheses

4v^2-15v-4=0

a = 4; b = -15; c = -4;
Δ = b2-4ac
Δ = -152-4·4·(-4)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-17}{2*4}=\frac{-2}{8}
=-1/4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+17}{2*4}=\frac{32}{8}
=4
$