5. \( 4 x^{2}+24 x+16=5 \)

To solve the equation \( 4x^2 + 24x + 16 = 5 \), start by moving the 5 to the left side to set the equation to zero: \[ 4x^2 + 24x + 16 - 5 = 0 \] \[ 4x^2 + 24x + 11 = 0 \] Next, you can use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), where \( a = 4 \), \( b = 24 \), and \( c = 11 \). Plug these values into the formula for a quick result! Using the quadratic formula will give you two solutions for \( x \): \[ x = \frac{-24 \pm \sqrt{24^2 - 4 \cdot 4 \cdot 11}}{2 \cdot 4} \] \[ x = \frac{-24 \pm \sqrt{576 - 176}}{8} \] \[ x = \frac{-24 \pm \sqrt{400}}{8} \] \[ x = \frac{-24 \pm 20}{8} \] This results in two possible values for \( x \): \( -0.5 \) and \( -5.5 \). To check your answers, plug these \( x \) values back into the original equation to ensure they satisfy it. Have fun exploring! To prevent errors, be careful with signs when manipulating the equation. Always double-check that you've rendered the quadratic correctly—transposing terms can lead to missed factors! Also, remember to simplify the radical in the quadratic formula completely; sometimes, factoring can reveal simpler solutions too. Happy solving!