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2g^2+15g+13=0
a = 2; b = 15; c = +13;
Δ = b2-4ac
Δ = 152-4·2·13
Δ = 121
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{121}=11$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-11}{2*2}=\frac{-26}{4} =-6+1/2 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+11}{2*2}=\frac{-4}{4} =-1 $
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