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40a+15a^2=0
a = 15; b = 40; c = 0;
Δ = b2-4ac
Δ = 402-4·15·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1600}=40$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40}{2*15}=\frac{-80}{30} =-2+2/3 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40}{2*15}=\frac{0}{30} =0 $
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