4x²+40x+84=0

Q: 4x²+40x+84=0

To solve the quadratic equation $4x^2 + 40x + 84 = 0$, we can use the quadratic formula, which is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where $a$, $b$, and $c$ are coefficients from the quadratic equation $ax^2 + bx + c = 0$. In this case, $a = 4$, $b = 40$, and $c = 84$. First, calculate the discriminant ($\Delta$): \[ \Delta = b^2 - 4ac = 40^2 - 4 \cdot 4 \cdot 84 = 1600 - 1344 = 256 \] Since the discriminant is positive, we have two distinct real roots. Now, we can find the roots: \[ x = \frac{-40 \pm \sqrt{256}}{2 \cdot 4} = \frac{-40 \pm 16}{8} \] There are two solutions: 1. For the plus sign: \[ x_1 = \frac{-40 + 16}{8} = \frac{-24}{8} = -3 \] 2. For the minus sign: \[ x_2 = \frac{-40 - 16}{8} = \frac{-56}{8} = -7 \] Therefore, the solutions to the equation $4x^2 + 40x + 84 = 0$ are $x_1 = -3$ and $x_2 = -7$.

Question

4x²+40x+84=0

Accepted Answer

To solve the quadratic equation $4x^2 + 40x + 84 = 0$, we can use the quadratic formula, which is given by:

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

where $a$, $b$, and $c$ are coefficients from the quadratic equation $ax^2 + bx + c = 0$. In this case, $a = 4$, $b = 40$, and $c = 84$.

First, calculate the discriminant ($\Delta$):

$$
\Delta = b^2 - 4ac = 40^2 - 4 \cdot 4 \cdot 84 = 1600 - 1344 = 256
$$

Since the discriminant is positive, we have two distinct real roots. Now, we can find the roots:

$$
x = \frac{-40 \pm \sqrt{256}}{2 \cdot 4} = \frac{-40 \pm 16}{8}
$$

There are two solutions:

1. For the plus sign:
$$
x_1 = \frac{-40 + 16}{8} = \frac{-24}{8} = -3
$$

2. For the minus sign:
$$
x_2 = \frac{-40 - 16}{8} = \frac{-56}{8} = -7
$$

Therefore, the solutions to the equation $4x^2 + 40x + 84 = 0$ are $x_1 = -3$ and $x_2 = -7$.