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12p^2+13p=0
a = 12; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·12·0
Δ = 169
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{169}=13$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*12}=\frac{-26}{24} =-1+1/12 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*12}=\frac{0}{24} =0 $
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