Question

use the discriminant to determine the number of real solutions of 3x^2 + 6x - 42 = 0 (1 point)

Explanation:

Step1: Identify coefficients

For the quadratic equation $3x^{2}+6x - 42=0$, we have $a = 3$, $b = 6$, $c=-42$.

Step2: Calculate the discriminant

The discriminant formula is $\Delta=b^{2}-4ac$. Substitute the values: $\Delta=(6)^{2}-4\times3\times(-42)$.
First, calculate $(6)^{2}=36$. Then, $4\times3\times(-42)=-504$. So, $\Delta = 36-(-504)=36 + 504=540$.

Step3: Determine the number of real - solutions

Since $\Delta=540>0$, the quadratic equation has two real solutions.

Answer:

Two real solutions