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2z^2=98
We move all terms to the left:
2z^2-(98)=0
a = 2; b = 0; c = -98;
Δ = b2-4ac
Δ = 02-4·2·(-98)
Δ = 784
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{784}=28$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-28}{2*2}=\frac{-28}{4} =-7 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+28}{2*2}=\frac{28}{4} =7 $
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