4v^2+12v-13=0

Solution for 4v^2+12v-13=0 equation:

4v^2+12v-13=0

a = 4; b = 12; c = -13;
Δ = b2-4ac
Δ = 122-4·4·(-13)
Δ = 352
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:

$\sqrt{\Delta}=\sqrt{352}=\sqrt{16*22}=\sqrt{16}*\sqrt{22}=4\sqrt{22}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{22}}{2*4}=\frac{-12-4\sqrt{22}}{8}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{22}}{2*4}=\frac{-12+4\sqrt{22}}{8}
$