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40x^2+41x=0
a = 40; b = 41; c = 0;
Δ = b2-4ac
Δ = 412-4·40·0
Δ = 1681
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{1681}=41$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-41}{2*40}=\frac{-82}{80} =-1+1/40 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+41}{2*40}=\frac{0}{80} =0 $
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