To determine the number of real solutions for the quadratic equation \(3x^2 + 6x - 42 = 0\), we first identify the coefficients \(a\), \(b\), and \(c\):
- \(a = 3\)
- \(b = 6\)
- \(c = -42\)
Next, we use the discriminant formula, which is given by:
\[ D = b^2 - 4ac \]
Substituting in the values of \(a\), \(b\), and \(c\):
\[ D = 6^2 - 4(3)(-42) \]
Calculating \(b^2\):
\[ 6^2 = 36 \]
Now calculating \(4ac\):
\[ 4(3)(-42) = -504 \]
So, the discriminant becomes:
\[ D = 36 - (-504) = 36 + 504 = 540 \]
Now we analyze the value of the discriminant \(D\):
- If \(D > 0\), there are two distinct real solutions.
- If \(D = 0\), there is one real solution (a repeated solution).
- If \(D < 0\), there are no real solutions.
Since \(D = 540\) is greater than 0, we conclude that the equation \(3x^2 + 6x - 42 = 0\) has two distinct real solutions.