12(y+1)=4(4y-3)y=

Solution for 12(y+1)=4(4y-3)y= equation:

12(y+1)=4(4y-3)y=

We move all terms to the left:

12(y+1)-(4(4y-3)y)=0

We multiply parentheses

12y-(4(4y-3)y)+12=0


We calculate terms in parentheses: -(4(4y-3)y), so:

4(4y-3)y

We multiply parentheses

16y^2-12y

Back to the equation:

-(16y^2-12y)


We get rid of parentheses

-16y^2+12y+12y+12=0

We add all the numbers together, and all the variables

-16y^2+24y+12=0

a = -16; b = 24; c = +12;
Δ = b2-4ac
Δ = 242-4·(-16)·12
Δ = 1344
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:

$\sqrt{\Delta}=\sqrt{1344}=\sqrt{64*21}=\sqrt{64}*\sqrt{21}=8\sqrt{21}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{21}}{2*-16}=\frac{-24-8\sqrt{21}}{-32}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{21}}{2*-16}=\frac{-24+8\sqrt{21}}{-32}
$