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5g^2+2g-6=0

a = 5; b = 2; c = -6;

Δ = b^{2}-4ac

Δ = 2^{2}-4·5·(-6)

Δ = 124

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}$

The end solution:

$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{31}}{2*5}=\frac{-2-2\sqrt{31}}{10} $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{31}}{2*5}=\frac{-2+2\sqrt{31}}{10} $

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