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[tex]
\begin{array}{l}
-3n^2 + 147 = 0 \\
n = \pm \square
\end{array}
[/tex]


Sagot :

Certainly! Let's solve the quadratic equation \(-3n^2 + 147 = 0\) step by step.

1. Start with the given equation:
[tex]\[ -3n^2 + 147 = 0 \][/tex]

2. Move 147 to the other side of the equation to isolate the quadratic term:
[tex]\[ -3n^2 = -147 \][/tex]

3. Divide both sides of the equation by -3 to solve for \(n^2\):
[tex]\[ n^2 = \frac{-147}{-3} \][/tex]
[tex]\[ n^2 = 49 \][/tex]

4. Take the square root of both sides to solve for \(n\):
[tex]\[ n = \pm \sqrt{49} \][/tex]

5. Calculate the square root of 49:
[tex]\[ n = \pm 7 \][/tex]

Therefore, the solutions are:
[tex]\[ n = \pm 7 \][/tex]

In the box, you should write:
[tex]\[ 7 \][/tex]