(v-7)(v-7)=2v^2+4v+70

Solution for (v-7)(v-7)=2v^2+4v+70 equation:

(v-7)(v-7)=2v^2+4v+70

We move all terms to the left:

(v-7)(v-7)-(2v^2+4v+70)=0

We get rid of parentheses

-2v^2+(v-7)(v-7)-4v-70=0

We multiply parentheses ..

-2v^2+(+v^2-7v-7v+49)-4v-70=0

We get rid of parentheses

-2v^2+v^2-7v-7v-4v+49-70=0

We add all the numbers together, and all the variables

-1v^2-18v-21=0

a = -1; b = -18; c = -21;
Δ = b2-4ac
Δ = -182-4·(-1)·(-21)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-4\sqrt{15}}{2*-1}=\frac{18-4\sqrt{15}}{-2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+4\sqrt{15}}{2*-1}=\frac{18+4\sqrt{15}}{-2}
$