(12+4y)y=5

Solution for (12+4y)y=5 equation:

(12+4y)y=5

We move all terms to the left:

(12+4y)y-(5)=0

We add all the numbers together, and all the variables

(4y+12)y-5=0

We multiply parentheses

4y^2+12y-5=0

a = 4; b = 12; c = -5;
Δ = b2-4ac
Δ = 122-4·4·(-5)
Δ = 224
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:

$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{14}}{2*4}=\frac{-12-4\sqrt{14}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{14}}{2*4}=\frac{-12+4\sqrt{14}}{8}
$