Sure! Let's solve the equation
[tex]\[
y = \sqrt{-4y + 21}.
\][/tex]
### Step-by-Step Solution:
1. Square both sides:
To eliminate the square root, we square both sides of the equation.
[tex]\[
y^2 = (-4y + 21).
\][/tex]
2. Rearrange the equation:
Move all terms to one side to set the equation to zero.
[tex]\[
y^2 + 4y - 21 = 0.
\][/tex]
3. Solve the quadratic equation:
We will solve the quadratic equation [tex]\(y^2 + 4y - 21 = 0\)[/tex]. This can be solved using the quadratic formula
[tex]\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
\][/tex]
where [tex]\(a = 1\)[/tex], [tex]\(b = 4\)[/tex], and [tex]\(c = -21\)[/tex].
4. Substitute values into the quadratic formula:
[tex]\[
y = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-21)}}{2 \cdot 1}.
\][/tex]
Simplify under the square root:
[tex]\[
y = \frac{-4 \pm \sqrt{16 + 84}}{2}.
\][/tex]
[tex]\[
y = \frac{-4 \pm \sqrt{100}}{2}.
\][/tex]
5. Simplify the square root:
[tex]\[
y = \frac{-4 \pm 10}{2}.
\][/tex]
6. Find the possible solutions:
[tex]\[
y = \frac{-4 + 10}{2}, \quad \text{and} \quad y = \frac{-4 - 10}{2}.
\][/tex]
[tex]\[
y = \frac{6}{2} = 3, \quad \text{and} \quad y = \frac{-14}{2} = -7.
\][/tex]
7. Check for extraneous solutions:
Substitute [tex]\(y = 3\)[/tex] back into the original equation:
[tex]\[
3 = \sqrt{-4(3) + 21},
\][/tex]
[tex]\[
3 = \sqrt{-12 + 21},
\][/tex]
[tex]\[
3 = \sqrt{9},
\][/tex]
[tex]\[
3 = 3.
\][/tex]
This solution is valid.
Now substitute [tex]\(y = -7\)[/tex] back into the original equation:
[tex]\[
-7 = \sqrt{-4(-7) + 21},
\][/tex]
[tex]\[
-7 = \sqrt{28 + 21},
\][/tex]
[tex]\[
-7 = \sqrt{49},
\][/tex]
[tex]\[
-7 = 7.
\][/tex]
This is not valid because [tex]\(-7 \neq 7\)[/tex].
### Conclusion:
Only the solution [tex]\(y = 3\)[/tex] is valid. Therefore, the solution to the equation is:
[tex]\[
y = 3.
\][/tex]
