Question 6 Solve for z using the square root property. \[ \begin{array}{l} 6 z^{2}+68=1+5 \\ z=\frac{1}{\frac{1}{1}} \end{array} \] Entry Tipc Do not use a decimal approximation for square roots. To enter a number like \( 5 \sqrt{7} \), type 5 syri( (7). Rreview your answer before submitting? Furthere Tips wite your ancwers as a list, weth your ancwers separated by (a) comma(s). For example, if you get 4 and \( -\frac{2}{3} \) as your answers, then enter \( 4,-2 / 3 \) in the bou. Next Question

Solve the equation \( 6z^{2}+68=1+5 \). Solve the equation(The complex numbers system) by following steps: - step0: Solve using square roots: \(6z^{2}+68=1+5\) - step1: Add the numbers: \(6z^{2}+68=6\) - step2: Add or subtract both sides: \(6z^{2}=-62\) - step3: Divide both sides: \(\frac{6z^{2}}{6}=\frac{-62}{6}\) - step4: Divide the numbers: \(z^{2}=-\frac{31}{3}\) - step5: Simplify the expression: \(z=\pm \sqrt{-\frac{31}{3}}\) - step6: Simplify the expression: \(z=\pm \frac{\sqrt{93}}{3}i\) - step7: Separate into possible cases: \(\begin{align}&z=\frac{\sqrt{93}}{3}i\\&z=-\frac{\sqrt{93}}{3}i\end{align}\) - step8: Rewrite: \(z_{1}=-\frac{\sqrt{93}}{3}i,z_{2}=\frac{\sqrt{93}}{3}i\) - step9: Remove the complex number(s): \(\textrm{No real solution}\) The equation \(6z^{2}+68=1+5\) does not have a real solution. Therefore, there is no value of \(z\) that satisfies the equation.