We are asked to solve the below quadratic equation by completing the square;
[tex]4x^2-12x+9=0[/tex]Step 1: We've got to rearrange the equation to have the constant term( the term without any variable) on the right hand side of the equation;
[tex]4x^2_{}-12x=-9[/tex]Step 2: We need the coefficient of x^2 to be 1, so we need to factor out 4 on the left hand side of the equation and divide both sides of the equation by 4;
[tex]\begin{gathered} 4(x^2-3x)=-9 \\ x^2-3x=-\frac{9}{4} \end{gathered}[/tex]Step 3: To complete the square, we'll find 1/2 of the coefficient of x which is -3 and square it. Then we'll add to both sides of the equation;
[tex]\begin{gathered} x^2-3x+(-\frac{3}{2})^2=-\frac{9}{4}+(-\frac{3}{2})^2 \\ x^2-3x+\frac{9}{4}=-\frac{9}{4}+\frac{9}{4} \\ x^2-3x+\frac{9}{4}=0 \end{gathered}[/tex]