-10v^2-41v-4=0

Solution for -10v^2-41v-4=0 equation:

-10v^2-41v-4=0

a = -10; b = -41; c = -4;
Δ = b2-4ac
Δ = -412-4·(-10)·(-4)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-39}{2*-10}=\frac{2}{-20}
=-1/10
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+39}{2*-10}=\frac{80}{-20}
=-4
$