(y-3)(y+3)=2y^2-12y+14

Solution for (y-3)(y+3)=2y^2-12y+14 equation:

(y-3)(y+3)=2y^2-12y+14

We move all terms to the left:

(y-3)(y+3)-(2y^2-12y+14)=0

We use the square of the difference formula

y^2-(2y^2-12y+14)-9=0

We get rid of parentheses

y^2-2y^2+12y-14-9=0

We add all the numbers together, and all the variables

-1y^2+12y-23=0

a = -1; b = 12; c = -23;
Δ = b2-4ac
Δ = 122-4·(-1)·(-23)
Δ = 52
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{13}}{2*-1}=\frac{-12-2\sqrt{13}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{13}}{2*-1}=\frac{-12+2\sqrt{13}}{-2}
$