4v^2-2v+1=25-2v

Solution for 4v^2-2v+1=25-2v equation:

4v^2-2v+1=25-2v

We move all terms to the left:

4v^2-2v+1-(25-2v)=0

We add all the numbers together, and all the variables

4v^2-2v-(-2v+25)+1=0

We get rid of parentheses

4v^2-2v+2v-25+1=0

We add all the numbers together, and all the variables

4v^2-24=0

a = 4; b = 0; c = -24;
Δ = b2-4ac
Δ = 02-4·4·(-24)
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{6}}{2*4}=\frac{0-8\sqrt{6}}{8}
=-\frac{8\sqrt{6}}{8}
=-\sqrt{6}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{6}}{2*4}=\frac{0+8\sqrt{6}}{8}
=\frac{8\sqrt{6}}{8}
=\sqrt{6}
$