2y2-32=0

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Solution for 2y2-32=0 equation: 



2y^2-32=0

a = 2; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·2·(-32)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*2}=\frac{-16}{4}
=-4
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*2}=\frac{16}{4}
=4
$

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