11y^2+48y-9=0

Solution for 11y^2+48y-9=0 equation:

11y^2+48y-9=0

a = 11; b = 48; c = -9;
Δ = b2-4ac
Δ = 482-4·11·(-9)
Δ = 2700
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:

$\sqrt{\Delta}=\sqrt{2700}=\sqrt{900*3}=\sqrt{900}*\sqrt{3}=30\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-30\sqrt{3}}{2*11}=\frac{-48-30\sqrt{3}}{22}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+30\sqrt{3}}{2*11}=\frac{-48+30\sqrt{3}}{22}
$