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-2.5x^2+3.5x=0
a = -2.5; b = 3.5; c = 0;
Δ = b2-4ac
Δ = 3.52-4·(-2.5)·0
Δ = 12.25
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{-(3.5)-\sqrt{12.25}}{2*-2.5}=\frac{-3.5-\sqrt{12.25}}{-5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.5)+\sqrt{12.25}}{2*-2.5}=\frac{-3.5+\sqrt{12.25}}{-5} $
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