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10x^2+18x=0
a = 10; b = 18; c = 0;
Δ = b2-4ac
Δ = 182-4·10·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{324}=18$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-18}{2*10}=\frac{-36}{20} =-1+4/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+18}{2*10}=\frac{0}{20} =0 $
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