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-0.5x^2+1.5x=0
a = -0.5; b = 1.5; c = 0;
Δ = b2-4ac
Δ = 1.52-4·(-0.5)·0
Δ = 2.25
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1.5)-\sqrt{2.25}}{2*-0.5}=\frac{-1.5-\sqrt{2.25}}{-1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.5)+\sqrt{2.25}}{2*-0.5}=\frac{-1.5+\sqrt{2.25}}{-1} $
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