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2z^2-288=0
a = 2; b = 0; c = -288;
Δ = b2-4ac
Δ = 02-4·2·(-288)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2304}=48$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*2}=\frac{-48}{4} =-12 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*2}=\frac{48}{4} =12 $
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