Answer
The graph of $f(x)=4x^2+3x-6$ has two $x-$intercepts.
Work Step by Step
The given function is
$\Rightarrow f(x)=4x^2+3x-6$
Find the number of real solutions of $0=4x^2+3x-6$
$= b^2-4ac$
Substitute $4$ for $a,3$ for $b,$ and $-6$ for $c$.
$= (3)^2-4(4)(-6)$
Simplify.
$=9+96$
Subtract.
$= 105$
Discriminant is greater than $0$,
Hence, the equation has two real solutions.
The graph of $f(x)=4x^2+3x-6$ has two $x-$intercepts.
