(2y+4)*(y+4)=196

Solution for (2y+4)*(y+4)=196 equation:

(2y+4)(y+4)=196

We move all terms to the left:

(2y+4)(y+4)-(196)=0

We multiply parentheses ..

(+2y^2+8y+4y+16)-196=0

We get rid of parentheses

2y^2+8y+4y+16-196=0

We add all the numbers together, and all the variables

2y^2+12y-180=0

a = 2; b = 12; c = -180;
Δ = b2-4ac
Δ = 122-4·2·(-180)
Δ = 1584
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:

$\sqrt{\Delta}=\sqrt{1584}=\sqrt{144*11}=\sqrt{144}*\sqrt{11}=12\sqrt{11}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{11}}{2*2}=\frac{-12-12\sqrt{11}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{11}}{2*2}=\frac{-12+12\sqrt{11}}{4}
$