2x2-16+8=0

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Solution for 2x2-16+8=0 equation:



2x^2-16+8=0
We add all the numbers together, and all the variables
2x^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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*2}=\frac{-8}{4} =-2 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*2}=\frac{8}{4} =2 $

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