### Instruction: Solve the inequality
\[-4x^2 + 7x + 2 < 0.\].### Response: To solve a quadratic inequality, I want to find the values of x that make the expression negative.
To do that, I need to find the zeros of the expression, or the values of x that make it equal to zero.
I can use the quadratic formula to find the zeros of the expression.
The formula is
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\]
where a, b, and c are the coefficients of the quadratic expression.
In this case, a = -4, b = 7, and c = 2.
Plugging these values into the formula, I get
\[x = \frac{-7 \pm \sqrt{7^2 - 4(-4)(2)}}{2(-4)}.\]
Simplifying the expression under the square root, I get
\[x = \frac{-7 \pm \sqrt{49 + 32}}{-8}.\]
Further simplifying, I get
\[x = \frac{-7 \pm \sqrt{81}}{-8}.\]
The square root of 81 is 9, so I can write
\[x = \frac{-7 \pm 9}{-8}.\]
This gives me two possible values for x: one when I add 9 and one when I subtract 9.
They are
\[x = \frac{-7 + 9}{-8} = \frac{2}{-8} = -\frac{1}{4}\]
and
\[x = \frac{-7 - 9}{-8} = \frac{-16}{-8} = 2.\]
These are the zeros of the expression, or the boundary points of the intervals where the expression changes sign.
To find out which intervals make the expression negative, I can test a sample value from each interval and see what sign the expression has for that value.
The intervals are (-infinity, -1/4), (-1/4, 2), and (2, infinity).
For the first interval, I can choose x = -1 as a sample value.
Plugging it into the expression, I get
\[-4(-1)^2 + 7(-1) + 2 = -4 - 7 + 2 = -9.\]
This is negative, so the expression is negative for all values of x in the interval (-infinity, -1/4).
For the second interval, I can choose x = 0 as a sample value.
Plugging it into the expression, I get
\[-4(0)^2 + 7(0) + 2 = 0 + 0 + 2 = 2.\]
This is positive, so the expression is positive for all values of x in the interval (-1/4, 2).
For the third interval, I can choose x = 3 as a sample value.
Plugging it into the expression, I get
\[-4(3)^2 + 7(3) + 2 = -36 + 21 + 2 = -13.\]
This is negative, so the expression is negative for all values of x in the interval (2, infinity).
Therefore, the solution to the inequality is the union of the intervals where the expression is negative, or
\[x \in (-\infty, -\frac{1}{4}) \cup (2, \infty).\]