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2g^2-15g+7=0
a = 2; b = -15; c = +7;
Δ = b2-4ac
Δ = -152-4·2·7
Δ = 169
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{169}=13$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-13}{2*2}=\frac{2}{4} =1/2 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+13}{2*2}=\frac{28}{4} =7 $
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