Question

Solve. (6x^2)(2x+3)=4(2x+3)+5x(2x+3)

Answer 1

We start with the equation:

\[
(6x^2)(2x + 3) = 4(2x + 3) + 5x(2x + 3)
\]

First, notice that both sides of the equation share the common factor \( (2x + 3) \). We can factor this out:

\[
(2x + 3)(6x^2) = (2x + 3)(4 + 5x)
\]

Assuming \( 2x + 3 \neq 0 \), we can divide both sides by \( 2x + 3 \):

\[
6x^2 = 4 + 5x
\]

Now, we will rearrange the equation to set it to 0:

\[
6x^2 - 5x - 4 = 0
\]

Next, we can solve this quadratic equation using the quadratic formula, where \( a = 6 \), \( b = -5 \), and \( c = -4 \):

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

Calculating the discriminant:

\[
b^2 - 4ac = (-5)^2 - 4(6)(-4)
\]
\[
= 25 + 96
\]
\[
= 121
\]

Now, we can apply the quadratic formula:

\[
x = \frac{-(-5) \pm \sqrt{121}}{2(6)}
\]
\[
= \frac{5 \pm 11}{12}
\]

Calculating the two possible solutions:

1.
\[
x = \frac{5 + 11}{12} = \frac{16}{12} = \frac{4}{3}
\]

2.
\[
x = \frac{5 - 11}{12} = \frac{-6}{12} = -\frac{1}{2}
\]

Thus, the solutions to the equation are:

\[
\boxed{\left\{ \frac{4}{3}, -\frac{1}{2} \right\}}
\]

Finally, we should check if \( 2x + 3 = 0 \) gives any additional solutions. Solving \( 2x + 3 = 0 \) gives:

\[
2x = -3 \quad \Rightarrow \quad x = -\frac{3}{2}
\]

Thus, we have an additional solution:

\[
x = -\frac{3}{2}
\]

The final set of solutions is therefore:

\[
\boxed{\left\{ \frac{4}{3}, -\frac{1}{2}, -\frac{3}{2} \right\}}
\]