2g^2+3g-5=0

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Solution for 2g^2+3g-5=0 equation: 



2g^2+3g-5=0

a = 2; b = 3; c = -5;
Δ = b2-4ac
Δ = 32-4·2·(-5)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$


$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-7}{2*2}=\frac{-10}{4}
=-2+1/2
$


$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*2}=\frac{4}{4}
=1
$

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