-4v^2-16v-15=0

Solution for -4v^2-16v-15=0 equation:

-4v^2-16v-15=0

a = -4; b = -16; c = -15;
Δ = b2-4ac
Δ = -162-4·(-4)·(-15)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4}{2*-4}=\frac{12}{-8}
=-1+1/2
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4}{2*-4}=\frac{20}{-8}
=-2+1/2
$