2y2-3y-5=0

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Solution for 2y2-3y-5=0 equation: 



2y^2-3y-5=0

a = 2; b = -3; c = -5;
Δ = b2-4ac
Δ = -32-4·2·(-5)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*2}=\frac{-4}{4}
=-1
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*2}=\frac{10}{4}
=2+1/2
$

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