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-0.5x^2+3x+5=0
a = -0.5; b = 3; c = +5;
Δ = b2-4ac
Δ = 32-4·(-0.5)·5
Δ = 19
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)-\sqrt{19}}{2*-0.5}=\frac{-3-\sqrt{19}}{-1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{19}}{2*-0.5}=\frac{-3+\sqrt{19}}{-1} $
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