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0.25q^2-6q+20=0
a = 0.25; b = -6; c = +20;
Δ = b2-4ac
Δ = -62-4·0.25·20
Δ = 16
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{16}=4$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-4}{2*0.25}=\frac{2}{0.5} =4 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4}{2*0.25}=\frac{10}{0.5} =20 $
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