-2g2-3g+7=0

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Solution for -2g2-3g+7=0 equation: 



-2g^2-3g+7=0

a = -2; b = -3; c = +7;
Δ = b2-4ac
Δ = -32-4·(-2)·7
Δ = 65
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{-(-3)-\sqrt{65}}{2*-2}=\frac{3-\sqrt{65}}{-4}
$


$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{65}}{2*-2}=\frac{3+\sqrt{65}}{-4}
$

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