16y-8y(3y-2)=-24

Solution for 16y-8y(3y-2)=-24 equation:

16y-8y(3y-2)=-24

We move all terms to the left:

16y-8y(3y-2)-(-24)=0

We add all the numbers together, and all the variables

16y-8y(3y-2)+24=0

We multiply parentheses

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

We add all the numbers together, and all the variables

-24y^2+32y+24=0

a = -24; b = 32; c = +24;
Δ = b2-4ac
Δ = 322-4·(-24)·24
Δ = 3328
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:

$\sqrt{\Delta}=\sqrt{3328}=\sqrt{256*13}=\sqrt{256}*\sqrt{13}=16\sqrt{13}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16\sqrt{13}}{2*-24}=\frac{-32-16\sqrt{13}}{-48}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16\sqrt{13}}{2*-24}=\frac{-32+16\sqrt{13}}{-48}
$