Show how many solutions are in the solution set for the equation 3x2=21 .(1 point) Responses two two one one zero zero infinitely many infinitely many

To solve the equation \(3x^2 = 21\), we first divide both sides by 3:

\[ x^2 = \frac{21}{3} = 7 \]

Next, we take the square root of both sides:

\[ x = \pm \sqrt{7} \]

This gives us two solutions:

1. \(x = \sqrt{7}\)
2. \(x = -\sqrt{7}\)

Therefore, there are two solutions in the solution set for the equation \(3x^2 = 21\).

Show how many solutions are in the solution set for 2(x−5)2+2=20 .(1 point) Responses infinitely many infinitely many zero zero one one two

Let's solve the equation \(2(x - 5)^2 + 2 = 20\).

First, we can subtract 2 from both sides:

\[ 2(x - 5)^2 = 20 - 2 \] \[ 2(x - 5)^2 = 18 \]

Next, divide both sides by 2:

\[ (x - 5)^2 = \frac{18}{2} = 9 \]

Now we take the square root of both sides:

\[ x - 5 = \pm 3 \]

This gives us two cases to solve for \(x\):

1. \(x - 5 = 3\): \[ x = 3 + 5 = 8 \]

2. \(x - 5 = -3\): \[ x = -3 + 5 = 2 \]

Thus, the solutions are \(x = 8\) and \(x = 2\). Therefore, there are two solutions in the solution set for the equation \(2(x - 5)^2 + 2 = 20\).