4y(2y-7)=11

Solution for 4y(2y-7)=11 equation:

4y(2y-7)=11

We move all terms to the left:

4y(2y-7)-(11)=0

We multiply parentheses

8y^2-28y-11=0

a = 8; b = -28; c = -11;
Δ = b2-4ac
Δ = -282-4·8·(-11)
Δ = 1136
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:

$\sqrt{\Delta}=\sqrt{1136}=\sqrt{16*71}=\sqrt{16}*\sqrt{71}=4\sqrt{71}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{71}}{2*8}=\frac{28-4\sqrt{71}}{16}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{71}}{2*8}=\frac{28+4\sqrt{71}}{16}
$