Question

Solve by using the Quadratic Formula:$3y^2-5y+2=0$

Short answer

The solution of the above equation is$y=1$and$y=\frac{2}{3}$

Step-by-step solution

Given information

We are given an equation$3y^2-5y+2=0$

Identify the values of a,b,c

On comparing with standard form we get the values as$a=3,b=-5,c=2$

Write the quadratic equation formula and substitute the values of a,b,c and simplify and solve for y

$y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substituting the values of $a,b,c$we get,

$$\begin{gathered} y=\frac{-(-5)\pm \sqrt{25-4\left(3\right)\left(2\right)}}{2\left(3\right)} \\ y=\frac{5\pm \sqrt{25-24}}{6} \\ y=\frac{5\pm 1}{6} \end{gathered}$$

on further simplifying

$$\begin{gathered} y=\frac{5+1}{6} \\ y=\frac{6}{6} \\ y=1 \end{gathered}$$and$$\begin{gathered} y=\frac{5-1}{6} \\ y=\frac{4}{6} \\ y=\frac{2}{3} \end{gathered}$$

Check the solutions

Now we substitute $y=1$
in the quadratic equation

$$\begin{gathered} 3y^2-5y+2 \\ 3{\left(1\right)}^2-5\left(1\right)+2 \\ 3-5+2 \\ =0 \end{gathered}$$

and now we substitute $x=\frac{2}{3}$in the equation, We get

$$\begin{gathered} 3y^2-5y+2 \\ 3{\left(\frac{2}{3}\right)}^2-5\left(\frac{2}{3}\right)+2 \\ \frac{4}{3}-\frac{10}{3}+2 \\ =0 \end{gathered}$$

Conclusion

The roots of the equation is$y=1$and $y=\frac{2}{3}$