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g^2-9g-10=0
a = 1; b = -9; c = -10;
Δ = b2-4ac
Δ = -92-4·1·(-10)
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$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-11}{2*1}=\frac{-2}{2} =-1 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+11}{2*1}=\frac{20}{2} =10 $
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