Questions: Find the real solution(s) of the given equation. (3 x+4)^2+2(3 x+4)-3=0 Give exact answers using fractions and square roots, not decimals. If there are multiple solutions, separate them with commas. If the function does not have a solution, enter DNE.

Solution Steps

To solve the given equation $(3x+4)^{2} + 2(3x+4) - 3 = 0$, we can use substitution to simplify it into a standard quadratic form. Let $u = 3x + 4$. The equation then becomes $u^2 + 2u - 3 = 0$. We can solve this quadratic equation for $u$ using the quadratic formula. Once we find the values of $u$, we substitute back to find the corresponding values of $x$.

We start with the equation $(3x+4)^{2} + 2(3x+4) - 3 = 0$. To simplify, let $u = 3x + 4$. The equation becomes: $$u^2 + 2u - 3 = 0$$

The equation $u^2 + 2u - 3 = 0$ is a standard quadratic equation. We solve it using the quadratic formula: $$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a = 1$, $b = 2$, and $c = -3$. Substituting these values, we get: $$u = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 12}}{2} = \frac{-2 \pm \sqrt{16}}{2}$$ $$u = \frac{-2 \pm 4}{2}$$ This gives us the solutions $u = 1$ and $u = -3$.

Now, we substitute back to find $x$ using $u = 3x + 4$.

For $u = 1$: $$1 = 3x + 4 \implies 3x = 1 - 4 \implies 3x = -3 \implies x = -1$$

For $u = -3$: $$-3 = 3x + 4 \implies 3x = -3 - 4 \implies 3x = -7 \implies x = -\frac{7}{3}$$

Final Answer