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0.4q^2+3q=0
a = 0.4; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·0.4·0
Δ = 9
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{9}=3$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*0.4}=\frac{-6}{0.8} =-7+0.4/0.8 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*0.4}=\frac{0}{0.8} =0 $
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