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6z^2=16
We move all terms to the left:
6z^2-(16)=0
a = 6; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·6·(-16)
Δ = 384
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{6}}{2*6}=\frac{0-8\sqrt{6}}{12} =-\frac{8\sqrt{6}}{12} =-\frac{2\sqrt{6}}{3} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{6}}{2*6}=\frac{0+8\sqrt{6}}{12} =\frac{8\sqrt{6}}{12} =\frac{2\sqrt{6}}{3} $
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