5y^2-4y-32=0

Solution for 5y^2-4y-32=0 equation:

5y^2-4y-32=0

a = 5; b = -4; c = -32;
Δ = b2-4ac
Δ = -42-4·5·(-32)
Δ = 656
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:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{41}}{2*5}=\frac{4-4\sqrt{41}}{10}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{41}}{2*5}=\frac{4+4\sqrt{41}}{10}
$