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0.5x^2+4.5x-14=0
a = 0.5; b = 4.5; c = -14;
Δ = b2-4ac
Δ = 4.52-4·0.5·(-14)
Δ = 48.25
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{-(4.5)-\sqrt{48.25}}{2*0.5}=\frac{-4.5-\sqrt{48.25}}{1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4.5)+\sqrt{48.25}}{2*0.5}=\frac{-4.5+\sqrt{48.25}}{1} $
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