2x+x^2-8=0

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Solution for 2x+x^2-8=0 equation: 



2x+x^2-8=0

a = 1; b = 2; c = -8;
Δ = b2-4ac
Δ = 22-4·1·(-8)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6}{2*1}=\frac{-8}{2}
=-4
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*1}=\frac{4}{2}
=2
$

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