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0.4q^2-18q+200=0
a = 0.4; b = -18; c = +200;
Δ = b2-4ac
Δ = -182-4·0.4·200
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4}=2$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2}{2*0.4}=\frac{16}{0.8} =20 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2}{2*0.4}=\frac{20}{0.8} =25 $
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