4y(3y+8)=113

Solution for 4y(3y+8)=113 equation:

4y(3y+8)=113

We move all terms to the left:

4y(3y+8)-(113)=0

We multiply parentheses

12y^2+32y-113=0

a = 12; b = 32; c = -113;
Δ = b2-4ac
Δ = 322-4·12·(-113)
Δ = 6448
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:

$\sqrt{\Delta}=\sqrt{6448}=\sqrt{16*403}=\sqrt{16}*\sqrt{403}=4\sqrt{403}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{403}}{2*12}=\frac{-32-4\sqrt{403}}{24}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{403}}{2*12}=\frac{-32+4\sqrt{403}}{24}
$