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2p^2-128=0
a = 2; b = 0; c = -128;
Δ = b2-4ac
Δ = 02-4·2·(-128)
Δ = 1024
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{1024}=32$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32}{2*2}=\frac{-32}{4} =-8 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32}{2*2}=\frac{32}{4} =8 $
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