4y^2-20y-96=0

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Solution for 4y^2-20y-96=0 equation:



4y^2-20y-96=0
a = 4; b = -20; c = -96;
Δ = b2-4ac
Δ = -202-4·4·(-96)
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-44}{2*4}=\frac{-24}{8} =-3 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+44}{2*4}=\frac{64}{8} =8 $

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