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z2=33
We move all terms to the left:
z2-(33)=0
We add all the numbers together, and all the variables
z^2-33=0
a = 1; b = 0; c = -33;
Δ = b2-4ac
Δ = 02-4·1·(-33)
Δ = 132
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{33}}{2*1}=\frac{0-2\sqrt{33}}{2} =-\frac{2\sqrt{33}}{2} =-\sqrt{33} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{33}}{2*1}=\frac{0+2\sqrt{33}}{2} =\frac{2\sqrt{33}}{2} =\sqrt{33} $
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