g^2+g-4=0

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Solution for g^2+g-4=0 equation: 



g^2+g-4=0

a = 1; b = 1; c = -4;
Δ = b2-4ac
Δ = 12-4·1·(-4)
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{17}}{2*1}=\frac{-1-\sqrt{17}}{2}
$


$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{17}}{2*1}=\frac{-1+\sqrt{17}}{2}
$

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