We are given the equation:

\[
62 = (4 + 16 + x)(5 + 8 + x)
\]

First, simplify the terms inside the parentheses:

\[
62 = (20 + x)(13 + x)
\]

Now, expand the expression on the right-hand side:

\[
62 = 20 \cdot 13 + 20 \cdot x + 13 \cdot x + x^2
\]

Simplify the expanded terms:

\[
62 = 260 + 33x + x^2
\]

Now, rearrange the equation to form a quadratic equation:

\[
x^2 + 33x + 260 = 62
\]

Subtract 62 from both sides:

\[
x^2 + 33x + 198 = 0
\]

Now, solve this quadratic equation. We can apply the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In this case, \(a = 1\), \(b = 33\), and \(c = 198\). Substitute these values into the quadratic formula:

\[
x = \frac{-33 \pm \sqrt{33^2 - 4(1)(198)}}{2(1)}
\]

First, calculate the discriminant:

\[
33^2 = 1089, \quad 4 \cdot 1 \cdot 198 = 792
\]
\[
\text{Discriminant} = 1089 - 792 = 297
\]

Now, calculate \(x\):

\[
x = \frac{-33 \pm \sqrt{297}}{2}
\]

Since \(\sqrt{297} \approx 17.23\), we get:

\[
x = \frac{-33 \pm 17.23}{2}
\]

So, the two solutions are:

\[
x_1 = \frac{-33 + 17.23}{2} = \frac{-15.77}{2} \approx -7.885
\]
\[
x_2 = \frac{-33 - 17.23}{2} = \frac{-50.23}{2} \approx -25.115
\]

Thus, the two approximate solutions for \(x\) are:

\[
x \approx -7.885 \quad \text{or} \quad x \approx -25.115
\]

Would you like more details or have any questions?

Here are five related questions:
1. How do you solve a quadratic equation by factoring?
2. What is the discriminant in a quadratic equation and how does it help?
3. Can you explain when and why the quadratic formula is used?
4. How do you determine if a quadratic equation has real or complex roots?
5. How can you approximate square roots manually?

**Tip:** Always check if a quadratic can be factored before using the quadratic formula; it can save time!

Learn how to solve the quadratic equation derived from 62 = (4 + 16 + x)(5 + 8 + x) with a step-by-step solution. This problem covers key algebraic concepts like expanding expressions and applying the quadratic formula. Suitable for students in Grades 9-12.

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