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18p^2-128=0
a = 18; b = 0; c = -128;
Δ = b2-4ac
Δ = 02-4·18·(-128)
Δ = 9216
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{9216}=96$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-96}{2*18}=\frac{-96}{36} =-2+2/3 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+96}{2*18}=\frac{96}{36} =2+2/3 $
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