2y^2+24y-5=0

Solution for 2y^2+24y-5=0 equation:

2y^2+24y-5=0

a = 2; b = 24; c = -5;
Δ = b2-4ac
Δ = 242-4·2·(-5)
Δ = 616
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:

$\sqrt{\Delta}=\sqrt{616}=\sqrt{4*154}=\sqrt{4}*\sqrt{154}=2\sqrt{154}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-2\sqrt{154}}{2*2}=\frac{-24-2\sqrt{154}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+2\sqrt{154}}{2*2}=\frac{-24+2\sqrt{154}}{4}
$