-4y-3=(12y-9)8y

Solution for -4y-3=(12y-9)8y equation:

-4y-3=(12y-9)8y

We move all terms to the left:

-4y-3-((12y-9)8y)=0


We calculate terms in parentheses: -((12y-9)8y), so:

(12y-9)8y

We multiply parentheses

96y^2-72y

Back to the equation:

-(96y^2-72y)


We get rid of parentheses

-96y^2-4y+72y-3=0

We add all the numbers together, and all the variables

-96y^2+68y-3=0

a = -96; b = 68; c = -3;
Δ = b2-4ac
Δ = 682-4·(-96)·(-3)
Δ = 3472
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:

$\sqrt{\Delta}=\sqrt{3472}=\sqrt{16*217}=\sqrt{16}*\sqrt{217}=4\sqrt{217}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(68)-4\sqrt{217}}{2*-96}=\frac{-68-4\sqrt{217}}{-192}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(68)+4\sqrt{217}}{2*-96}=\frac{-68+4\sqrt{217}}{-192}
$