3y^2+24y=28

Solution for 3y^2+24y=28 equation:

3y^2+24y=28

We move all terms to the left:

3y^2+24y-(28)=0

a = 3; b = 24; c = -28;
Δ = b2-4ac
Δ = 242-4·3·(-28)
Δ = 912
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:

$\sqrt{\Delta}=\sqrt{912}=\sqrt{16*57}=\sqrt{16}*\sqrt{57}=4\sqrt{57}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{57}}{2*3}=\frac{-24-4\sqrt{57}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{57}}{2*3}=\frac{-24+4\sqrt{57}}{6}
$