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2z^2=z^2+225
We move all terms to the left:
2z^2-(z^2+225)=0
We get rid of parentheses
2z^2-z^2-225=0
We add all the numbers together, and all the variables
z^2-225=0
a = 1; b = 0; c = -225;
Δ = b2-4ac
Δ = 02-4·1·(-225)
Δ = 900
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{900}=30$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-30}{2*1}=\frac{-30}{2} =-15 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+30}{2*1}=\frac{30}{2} =15 $
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