V^4-8v^2+v+5=0

Solution for V^4-8v^2+v+5=0 equation:

^4-8V^2+V+5=0

We add all the numbers together, and all the variables

-8V^2+V=0

a = -8; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-8)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-8}=\frac{-2}{-16}
=1/8
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-8}=\frac{0}{-16}
=0
$