4(2y+1)=2(12-y)y=

Solution for 4(2y+1)=2(12-y)y= equation:

4(2y+1)=2(12-y)y=

We move all terms to the left:

4(2y+1)-(2(12-y)y)=0

We add all the numbers together, and all the variables

4(2y+1)-(2(-1y+12)y)=0

We multiply parentheses

8y-(2(-1y+12)y)+4=0


We calculate terms in parentheses: -(2(-1y+12)y), so:

2(-1y+12)y

We multiply parentheses

-2y^2+24y

Back to the equation:

-(-2y^2+24y)


We get rid of parentheses

2y^2-24y+8y+4=0

We add all the numbers together, and all the variables

2y^2-16y+4=0

a = 2; b = -16; c = +4;
Δ = b2-4ac
Δ = -162-4·2·4
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{-(-16)-4\sqrt{14}}{2*2}=\frac{16-4\sqrt{14}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{14}}{2*2}=\frac{16+4\sqrt{14}}{4}
$