91Ó°ÊÓ

The given equation is

$\frac{4}{3}{y}^2−4y+3=0$

First, write the equation $a{x}^2+bx+c=0$a in standard form, as shown below.

To get rid of the fraction, multiply both sides of the equation by 3.

role="math" localid="1653832085826" $$\begin{gathered} 3\left(\frac{4}{3}{y}^2-4y+3\right)=0·3 \\ 3·\frac{4}{3}{y}^2-3·3y+3.3=0 \\ 4y^2-12y+9=0 \end{gathered}$$

In standard form, the equation is $4y^2-12y+9=0$

Determine the a, b, and c values.

In comparison to the standard form, the values are a = 4,b = -12, and c= 9.

In the quadratic formula, substitute a = 4, b = -12, and c= 9.

role="math" localid="1653831910289" $y=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a}$

Put the values.

role="math" localid="1653831896763" $y=\frac{-\left(-12\right)Â\pm \sqrt{{\left(-12\right)}^2-4\left(4\right)\left(9\right)}}{2\left(4\right)}$

As shown below, simplify the fraction.

role="math" localid="1653831941126" $$\begin{gathered} y=\frac{-\left(-12\right)Â\pm \sqrt{{\left(-12\right)}^2-4\left(4\right)\left(9\right)}}{2\left(4\right)} \\ y=\frac{12Â\pm \sqrt{144+-144}}{8} \\ y=\frac{12Â\pm \sqrt{0}}{8} \\ y=\frac{12}{8} \\ y=\frac{3}{2} \end{gathered}$$

As a result is $y=\frac{3}{2}$

To test the solutions, enter $y=\frac{3}{2}$into the original equation.

$$\begin{gathered} 4y^2-12y+9=0 \\ \frac{4}{3}{\left(\frac{3}{2}\right)}^2-4\left(\frac{3}{2}\right)+3=0 \\ 3-6+3=0 \\ 0=0 \end{gathered}$$

Therefore, the solution is$y=\frac{3}{2}$