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18p+45p^2=0
a = 45; b = 18; c = 0;
Δ = b2-4ac
Δ = 182-4·45·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{324}=18$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-18}{2*45}=\frac{-36}{90} =-2/5 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+18}{2*45}=\frac{0}{90} =0 $
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