b^2+b-8=0

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



b^2+b-8=0

a = 1; b = 1; c = -8;
Δ = b2-4ac
Δ = 12-4·1·(-8)
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{33}}{2*1}=\frac{-1-\sqrt{33}}{2}
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{33}}{2*1}=\frac{-1+\sqrt{33}}{2}
$

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