2y^2+0.5y-6=0

Solution for 2y^2+0.5y-6=0 equation:

2y^2+0.5y-6=0

a = 2; b = 0.5; c = -6;
Δ = b2-4ac
Δ = 0.52-4·2·(-6)
Δ = 48.25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.5)-\sqrt{48.25}}{2*2}=\frac{-0.5-\sqrt{48.25}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{48.25}}{2*2}=\frac{-0.5+\sqrt{48.25}}{4}
$