4v^2+v^2=128

Solution for 4v^2+v^2=128 equation:

4v^2+v^2=128

We move all terms to the left:

4v^2+v^2-(128)=0

We add all the numbers together, and all the variables

5v^2-128=0

a = 5; b = 0; c = -128;
Δ = b2-4ac
Δ = 02-4·5·(-128)
Δ = 2560
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:

$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{10}}{2*5}=\frac{0-16\sqrt{10}}{10}
=-\frac{16\sqrt{10}}{10}
=-\frac{8\sqrt{10}}{5}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{10}}{2*5}=\frac{0+16\sqrt{10}}{10}
=\frac{16\sqrt{10}}{10}
=\frac{8\sqrt{10}}{5}
$