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4z^2-196=0
a = 4; b = 0; c = -196;
Δ = b2-4ac
Δ = 02-4·4·(-196)
Δ = 3136
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{3136}=56$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-56}{2*4}=\frac{-56}{8} =-7 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+56}{2*4}=\frac{56}{8} =7 $
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