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0.04q^2-1.4q+3=0
a = 0.04; b = -1.4; c = +3;
Δ = b2-4ac
Δ = -1.42-4·0.04·3
Δ = 1.48
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}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1.4)-\sqrt{1.48}}{2*0.04}=\frac{1.4-\sqrt{1.48}}{0.08} $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.4)+\sqrt{1.48}}{2*0.04}=\frac{1.4+\sqrt{1.48}}{0.08} $
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