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2z^2-5=13
We move all terms to the left:
2z^2-5-(13)=0
We add all the numbers together, and all the variables
2z^2-18=0
a = 2; b = 0; c = -18;
Δ = b2-4ac
Δ = 02-4·2·(-18)
Δ = 144
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{144}=12$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*2}=\frac{-12}{4} =-3 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*2}=\frac{12}{4} =3 $
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