4y^2+10y-65=0

Solution for 4y^2+10y-65=0 equation:

4y^2+10y-65=0

a = 4; b = 10; c = -65;
Δ = b2-4ac
Δ = 102-4·4·(-65)
Δ = 1140
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:

$\sqrt{\Delta}=\sqrt{1140}=\sqrt{4*285}=\sqrt{4}*\sqrt{285}=2\sqrt{285}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{285}}{2*4}=\frac{-10-2\sqrt{285}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{285}}{2*4}=\frac{-10+2\sqrt{285}}{8}
$