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z^2=0.6
We move all terms to the left:
z^2-(0.6)=0
We add all the numbers together, and all the variables
z^2-0.6=0
a = 1; b = 0; c = -0.6;
Δ = b2-4ac
Δ = 02-4·1·(-0.6)
Δ = 2.4
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{2.4}}{2*1}=\frac{0-\sqrt{2.4}}{2} =-\frac{\sqrt{}}{2} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{2.4}}{2*1}=\frac{0+\sqrt{2.4}}{2} =\frac{\sqrt{}}{2} $
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