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0.5x^2+x-1.5=0
a = 0.5; b = 1; c = -1.5;
Δ = b2-4ac
Δ = 12-4·0.5·(-1.5)
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-2}{2*0.5}=\frac{-3}{1} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+2}{2*0.5}=\frac{1}{1} =1 $
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