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0.5x^2+x-4=0
a = 0.5; b = 1; c = -4;
Δ = b2-4ac
Δ = 12-4·0.5·(-4)
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3}{2*0.5}=\frac{-4}{1} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3}{2*0.5}=\frac{2}{1} =2 $
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