# Question #8d622

May 8, 2017

$0$

#### Explanation:

We have:

$f \left(x\right) = 3 {x}^{2} + 3 x + 2$

An $x$-intercept occurs when $f \left(x\right) = 0$

$\implies 3 {x}^{2} + 3 x + 2 = 0$

To solve this quadratic we could compete the square or use the quadratic formula,

We can also use the discriminant $\Delta = {b}^{2} - 4 a c$ to determine the nature of the roots of $a {x}^{2} + b x + c = 0$, as:

If $\Delta = {b}^{2} - 4 a c \setminus \left\{\begin{matrix}< 0 & \text{no real roots" \\ =0 & "two equal real roots" \\ gt 0 & "two distinct real roots}\end{matrix}\right.$

For our equation we have:

$\Delta = 3 \cdot 3 - 4 \cdot 2 \cdot 2 < 0$

Hence there are no real solutions, and therefore $0$ $x$-intercepts.

We can confirm this graphically:
graph{3x^2+3x+2 [-5, 5, -2, 5]}