12y^2-28y=15

Solution for 12y^2-28y=15 equation:

12y^2-28y=15

We move all terms to the left:

12y^2-28y-(15)=0

a = 12; b = -28; c = -15;
Δ = b2-4ac
Δ = -282-4·12·(-15)
Δ = 1504
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:

$\sqrt{\Delta}=\sqrt{1504}=\sqrt{16*94}=\sqrt{16}*\sqrt{94}=4\sqrt{94}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{94}}{2*12}=\frac{28-4\sqrt{94}}{24}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{94}}{2*12}=\frac{28+4\sqrt{94}}{24}
$