3(y+4)4y=15

Solution for 3(y+4)4y=15 equation:

3(y+4)4y=15

We move all terms to the left:

3(y+4)4y-(15)=0

We multiply parentheses

12y^2+48y-15=0

a = 12; b = 48; c = -15;
Δ = b2-4ac
Δ = 482-4·12·(-15)
Δ = 3024
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:

$\sqrt{\Delta}=\sqrt{3024}=\sqrt{144*21}=\sqrt{144}*\sqrt{21}=12\sqrt{21}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-12\sqrt{21}}{2*12}=\frac{-48-12\sqrt{21}}{24}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+12\sqrt{21}}{2*12}=\frac{-48+12\sqrt{21}}{24}
$