The problem asks us to solve the quadratic equation $4x^2 + 20x + 21 = 0$ using the quadratic formula and give any decimal answers to 1 decimal place.

1. Problem Description

The problem asks us to solve the quadratic equation $4x^2 + 20x + 21 = 0$ using the quadratic formula and give any decimal answers to 1 decimal place.

2. Solution Steps

The quadratic formula is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the given equation $4x^2 + 20x + 21 = 0$, we have $a = 4$, $b = 20$, and $c = 21$.

Substituting these values into the quadratic formula, we get:

$x = \frac{-20 \pm \sqrt{20^2 - 4(4)(21)}}{2(4)}$

$x = \frac{-20 \pm \sqrt{400 - 336}}{8}$

$x = \frac{-20 \pm \sqrt{64}}{8}$

$x = \frac{-20 \pm 8}{8}$

Now, we have two possible solutions for $x$:

$x_1 = \frac{-20 + 8}{8} = \frac{-12}{8} = -\frac{3}{2} = -1.5$

$x_2 = \frac{-20 - 8}{8} = \frac{-28}{8} = -\frac{7}{2} = -3.5$

3. Final Answer

The two possible values of $x$ are -1.5 and -3.

5. Final Answer: $x = -1.5, -3.5$