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