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16p^2+15p=0
a = 16; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·16·0
Δ = 225
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{225}=15$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15}{2*16}=\frac{-30}{32} =-15/16 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*16}=\frac{0}{32} =0 $
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