4v^2=-13v+3

Solution for 4v^2=-13v+3 equation:

4v^2=-13v+3

We move all terms to the left:

4v^2-(-13v+3)=0

We get rid of parentheses

4v^2+13v-3=0

a = 4; b = 13; c = -3;
Δ = b2-4ac
Δ = 132-4·4·(-3)
Δ = 217
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{217}}{2*4}=\frac{-13-\sqrt{217}}{8}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{217}}{2*4}=\frac{-13+\sqrt{217}}{8}
$