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z^2=50
We move all terms to the left:
z^2-(50)=0
a = 1; b = 0; c = -50;
Δ = b2-4ac
Δ = 02-4·1·(-50)
Δ = 200
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{2}}{2*1}=\frac{0-10\sqrt{2}}{2} =-\frac{10\sqrt{2}}{2} =-5\sqrt{2} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{2}}{2*1}=\frac{0+10\sqrt{2}}{2} =\frac{10\sqrt{2}}{2} =5\sqrt{2} $
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