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16p^2-64=0
a = 16; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·16·(-64)
Δ = 4096
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{4096}=64$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-64}{2*16}=\frac{-64}{32} =-2 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+64}{2*16}=\frac{64}{32} =2 $
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