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2z^2-24=0
a = 2; b = 0; c = -24;
Δ = b2-4ac
Δ = 02-4·2·(-24)
Δ = 192
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*2}=\frac{0-8\sqrt{3}}{4} =-\frac{8\sqrt{3}}{4} =-2\sqrt{3} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*2}=\frac{0+8\sqrt{3}}{4} =\frac{8\sqrt{3}}{4} =2\sqrt{3} $
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