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3p^2-16p+16=0
a = 3; b = -16; c = +16;
Δ = b2-4ac
Δ = -162-4·3·16
Δ = 64
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{64}=8$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8}{2*3}=\frac{8}{6} =1+1/3 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8}{2*3}=\frac{24}{6} =4 $
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