4y^2-3y-5=0

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Solution for 4y^2-3y-5=0 equation: 



4y^2-3y-5=0

a = 4; b = -3; c = -5;
Δ = b2-4ac
Δ = -32-4·4·(-5)
Δ = 89
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{-(-3)-\sqrt{89}}{2*4}=\frac{3-\sqrt{89}}{8}
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{89}}{2*4}=\frac{3+\sqrt{89}}{8}
$

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