2(y+4)*(y+4)=22

Solution for 2(y+4)*(y+4)=22 equation:

2(y+4)(y+4)=22

We move all terms to the left:

2(y+4)(y+4)-(22)=0

We multiply parentheses ..

2(+y^2+4y+4y+16)-22=0

We multiply parentheses

2y^2+8y+8y+32-22=0

We add all the numbers together, and all the variables

2y^2+16y+10=0

a = 2; b = 16; c = +10;
Δ = b2-4ac
Δ = 162-4·2·10
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{11}}{2*2}=\frac{-16-4\sqrt{11}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{11}}{2*2}=\frac{-16+4\sqrt{11}}{4}
$