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40c^2+24c=0
a = 40; b = 24; c = 0;
Δ = b2-4ac
Δ = 242-4·40·0
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{576}=24$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-24}{2*40}=\frac{-48}{80} =-3/5 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24}{2*40}=\frac{0}{80} =0 $
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