4y^2-8y-32=0

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Solution for 4y^2-8y-32=0 equation:



4y^2-8y-32=0
a = 4; b = -8; c = -32;
Δ = b2-4ac
Δ = -82-4·4·(-32)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-24}{2*4}=\frac{-16}{8} =-2 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+24}{2*4}=\frac{32}{8} =4 $

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