4v^2+2v-7=0

Solution for 4v^2+2v-7=0 equation:

4v^2+2v-7=0

a = 4; b = 2; c = -7;
Δ = b2-4ac
Δ = 22-4·4·(-7)
Δ = 116
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:

$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{29}}{2*4}=\frac{-2-2\sqrt{29}}{8}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{29}}{2*4}=\frac{-2+2\sqrt{29}}{8}
$