(V-3)^2=2v^2-4v-6

Solution for (V-3)^2=2v^2-4v-6 equation:

(-3)^2=2V^2-4V-6

We move all terms to the left:

(-3)^2-(2V^2-4V-6)=0

We add all the numbers together, and all the variables

-(2V^2-4V-6)+9=0

We get rid of parentheses

-2V^2+4V+6+9=0

We add all the numbers together, and all the variables

-2V^2+4V+15=0

a = -2; b = 4; c = +15;
Δ = b2-4ac
Δ = 42-4·(-2)·15
Δ = 136
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:

$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{34}}{2*-2}=\frac{-4-2\sqrt{34}}{-4}
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{34}}{2*-2}=\frac{-4+2\sqrt{34}}{-4}
$