How do you solve \[{{y}^{2}}-3y+2=0\]?

Hint: We have to solve the problem using discriminant method. First we have to calculate the discriminant using the formula. Next you have to substitute it in the quadratic formula to get the factors of the equation and the results for variable \[y\].

Complete step by step answer:
Given equation is
\[{{y}^{2}}-3y+2=0\]
First we have to find the discriminant of the equation.
The discriminant for standard equation \[a{{x}^{2}}-bx+c=0\] is \[{{b}^{2}}-4ac\]
From this we can write \[a,b,c\] values for our equation they are
\[\begin{align}
  & a=1 \\
 & b=-3 \\
 & c=2 \\
\end{align}\]
Using these values we can find the discriminant of our equation
 \[{{b}^{2}}-4ac={{\left( -3 \right)}^{2}}-4\times 1\times 2\]
By simplifying it we get
\[\Rightarrow 9-8\]
\[\Rightarrow 1\]
So the discriminant value is \[1\].
Now we have substituted this discriminant value in the quadratic formula.
The quadratic formula for the standard equation is \[a{{x}^{2}}-bx+c=0\] is
\[y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Now we have to substitute the values we got from our equation.
We can observe the discriminant in the formula. So first we have to substitute the discriminant in the formula. Then we will get
\[\Rightarrow \dfrac{-b\pm \sqrt{1}}{2a}\]
 Now we have substituted all the values that we got from our equation in the formula. Then we will get
\[\Rightarrow \dfrac{-\left( -3 \right)\pm \sqrt{1}}{2\times 1}\]
Now we have to simplify the equation.
As we all know the square root of \[1\] is \[1\]. By substituting it we will get
\[\Rightarrow \dfrac{-\left( -3 \right)\pm 1}{2\times 1}\]
Here we have two negative signs for \[3\]. We can make it as positive because two negative signs products will give a positive number. Then the formula will be like
\[\Rightarrow \dfrac{3\pm 1}{2\times 1}\]
\[\Rightarrow \dfrac{3\pm 1}{2}\]
Now we can make them two values. One with adding the terms in numerator and other by subtracting the terms in numerator.
\[\begin{align}
  & \Rightarrow \dfrac{3+1}{2} \\
 & \Rightarrow \dfrac{3-1}{2} \\
\end{align}\]
We will get
\[\Rightarrow \dfrac{4}{2}\]
\[\Rightarrow \dfrac{2}{2}\]
So the values for \[y\] are \[2\] and \[1\] for the given equation \[{{y}^{2}}-3y+2=0\].
\[y=2,y=1\]

Note: We can also ask this question using factoring method. But by solving in this form we can know whether we will get real or imaginary roots. We can also check the values by substituting them in the equation whether they are correct or not.