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-10g^2+32g=0
a = -10; b = 32; c = 0;
Δ = b2-4ac
Δ = 322-4·(-10)·0
Δ = 1024
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{1024}=32g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32}{2*-10}=\frac{-64}{-20} =3+1/5g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32}{2*-10}=\frac{0}{-20} =0
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