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40x^2+15x=0
a = 40; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·40·0
Δ = 225
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{225}=15$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15}{2*40}=\frac{-30}{80} =-3/8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*40}=\frac{0}{80} =0 $
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