Use factoring to solve the quadratic equation. Check by substitution or by using a graphing utility and identifying $x$-intercepts.
\[
4 x^{2}=27 x+81
\]
Rewrite the equation in factored form.
$=0$
(Factor completely.)
The solution set is
(Use commas to separate answers as needed. Type each solution only once.)
Answer
Finally, we can check our solution by substituting \(x = \frac{9}{2}\) back into the original equation: \(4(\frac{9}{2})^2 = 27(\frac{9}{2}) + 81\), which simplifies to \(81 = 81\), confirming that our solution is correct.
Step 1 ：First, rewrite the equation in standard form: \(4x^2 - 27x - 81 = 0\).
Step 2 ：Next, factor the quadratic equation. To do this, we need to find two numbers that multiply to \(-4 \times -81 = 324\) and add to \(-27\). These numbers are \(-18\) and \(-9\).
Step 3 ：So, we can rewrite the equation as \((4x^2 - 18x) - (9x + 81) = 0\).
Step 4 ：Factor by grouping: \(2x(2x - 9) - 9(2x - 9) = 0\).
Step 5 ：This gives us \((2x - 9)(2x - 9) = 0\).
Step 6 ：Setting each factor equal to zero gives the solutions \(x = \frac{9}{2}\) and \(x = \frac{9}{2}\).
Step 7 ：So, the solution set is \(\boxed{\{\frac{9}{2}\}}\).
Step 8 ：Finally, we can check our solution by substituting \(x = \frac{9}{2}\) back into the original equation: \(4(\frac{9}{2})^2 = 27(\frac{9}{2}) + 81\), which simplifies to \(81 = 81\), confirming that our solution is correct.
