12y+48-4y=8y(y-6)

Solution for 12y+48-4y=8y(y-6) equation:

12y+48-4y=8y(y-6)

We move all terms to the left:

12y+48-4y-(8y(y-6))=0

We add all the numbers together, and all the variables

8y-(8y(y-6))+48=0


We calculate terms in parentheses: -(8y(y-6)), so:

8y(y-6)

We multiply parentheses

8y^2-48y

Back to the equation:

-(8y^2-48y)


We get rid of parentheses

-8y^2+8y+48y+48=0

We add all the numbers together, and all the variables

-8y^2+56y+48=0

a = -8; b = 56; c = +48;
Δ = b2-4ac
Δ = 562-4·(-8)·48
Δ = 4672
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:

$\sqrt{\Delta}=\sqrt{4672}=\sqrt{64*73}=\sqrt{64}*\sqrt{73}=8\sqrt{73}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-8\sqrt{73}}{2*-8}=\frac{-56-8\sqrt{73}}{-16}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+8\sqrt{73}}{2*-8}=\frac{-56+8\sqrt{73}}{-16}
$