2r2-8=0

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Solution for 2r2-8=0 equation: 



2r^2-8=0

a = 2; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·2·(-8)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*2}=\frac{-8}{4}
=-2
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*2}=\frac{8}{4}
=2
$

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