To find the solutions to the quadratic equation [tex]\( 3x^2 + 45x + 24 = 0 \)[/tex], we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the given quadratic equation, the coefficients are:
[tex]\[ a = 3, \quad b = 45, \quad c = 24 \][/tex]
First, we'll calculate the discriminant ([tex]\( \Delta \)[/tex]):
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = 45^2 - 4 \cdot 3 \cdot 24 \][/tex]
[tex]\[ \Delta = 2025 - 288 \][/tex]
[tex]\[ \Delta = 1737 \][/tex]
The quadratic formula provides us with two potential solutions:
[tex]\[ x = \frac{-b + \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ x = \frac{-b - \sqrt{\Delta}}{2a} \][/tex]
We are given that one of the solutions is [tex]\( x = -14.45 \)[/tex].
To find the other solution, we need to substitute the values into the formula. Let's consider both cases:
Case 1:
[tex]\[ x = \frac{-45 + \sqrt{1737}}{6} \][/tex]
Case 2:
[tex]\[ x = \frac{-45 - \sqrt{1737}}{6} \][/tex]
Given that one solution is [tex]\( x = -14.45 \)[/tex]:
Checking Case 1:
[tex]\[ -14.45 = \frac{-45 + \sqrt{1737}}{6} \][/tex]
Solving for [tex]\( \sqrt{1737} \)[/tex]:
[tex]\[ -14.45 \cdot 6 = -45 + \sqrt{1737} \][/tex]
[tex]\[ -86.7 = -45 + \sqrt{1737} \][/tex]
[tex]\[ \sqrt{1737} = -86.7 + 45 \][/tex]
[tex]\[ \sqrt{1737} = -41.7 \][/tex]
This does not provide a valid square root, so this case is incorrect.
Checking Case 2:
[tex]\[ x = \frac{-45 - \sqrt{1737}}{6} \][/tex]
Given [tex]\( -14.45 \neq \frac{-45 + \sqrt{1737}}{6} \)[/tex], it must be the following solution:
[tex]\[ -14.45 = \frac{-45 - \sqrt{1737}}{6} \][/tex]
To find the other solution:
[tex]\[ x = \frac{-45 + \sqrt{1737}}{6} \][/tex]
Calculate:
[tex]\[ x = \frac{-45 + \sqrt{1737}}{6} \][/tex]
[tex]\[ x = \frac{-45 + 41.7}{6} \][/tex]
[tex]\[ x = \frac{-3.3}{6} \][/tex]
[tex]\[ x \approx -0.55 \][/tex]
Thus, the other solution is approximately [tex]\(-0.55\)[/tex].
The answer is:
[tex]\[ \boxed{-0.55} \][/tex]
