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0.2x^2-9x+40=0
a = 0.2; b = -9; c = +40;
Δ = b2-4ac
Δ = -92-4·0.2·40
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-7}{2*0.2}=\frac{2}{0.4} =5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+7}{2*0.2}=\frac{16}{0.4} =40 $
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