Use the discriminant to tell whether the solutions of this equation are real or complex \[ 4 x^{2}-3 x+9=0 \]

Solve the equation \( 4x^2-3x+9=0 \). Solve the equation(The complex numbers system) by following steps: - step0: Solve using the quadratic formula in the complex numbers system: \(4x^{2}-3x+9=0\) - step1: Solve using the quadratic formula: \(x=\frac{3\pm \sqrt{\left(-3\right)^{2}-4\times 4\times 9}}{2\times 4}\) - step2: Simplify the expression: \(x=\frac{3\pm \sqrt{\left(-3\right)^{2}-4\times 4\times 9}}{8}\) - step3: Simplify the expression: \(x=\frac{3\pm \sqrt{-135}}{8}\) - step4: Simplify the expression: \(x=\frac{3\pm 3\sqrt{15}\times i}{8}\) - step5: Separate into possible cases: \(\begin{align}&x=\frac{3+3\sqrt{15}\times i}{8}\\&x=\frac{3-3\sqrt{15}\times i}{8}\end{align}\) - step6: Simplify the expression: \(\begin{align}&x=\frac{3}{8}+\frac{3\sqrt{15}}{8}i\\&x=\frac{3-3\sqrt{15}\times i}{8}\end{align}\) - step7: Simplify the expression: \(\begin{align}&x=\frac{3}{8}+\frac{3\sqrt{15}}{8}i\\&x=\frac{3}{8}-\frac{3\sqrt{15}}{8}i\end{align}\) - step8: Rewrite: \(x_{1}=\frac{3}{8}-\frac{3\sqrt{15}}{8}i,x_{2}=\frac{3}{8}+\frac{3\sqrt{15}}{8}i\) - step9: Remove the complex number(s): \(\textrm{No real solution}\) The solutions of the equation \(4x^{2}-3x+9=0\) are complex. Therefore, the solutions are not real.