Certainly! Let's solve the equation:
[tex]\[ x^2 + 4x = 3 \][/tex]
We will go through the steps to solve this quadratic equation.
### Step 1: Rewrite the equation in standard quadratic form
First, let's get all terms to one side of the equation to write it in the standard quadratic form [tex]\(ax^2 + bx + c = 0\)[/tex].
[tex]\[ x^2 + 4x - 3 = 0 \][/tex]
### Step 2: Identify the coefficients
In our quadratic equation [tex]\(x^2 + 4x - 3 = 0\)[/tex], the coefficients are:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = 4\)[/tex]
- [tex]\(c = -3\)[/tex]
### Step 3: Use the quadratic formula
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
### Step 4: Plug in the coefficients into the quadratic formula
Let's substitute [tex]\(a = 1\)[/tex], [tex]\(b = 4\)[/tex], and [tex]\(c = -3\)[/tex] into the quadratic formula.
[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \][/tex]
### Step 5: Simplify under the square root
Calculate the discriminant (inside the square root):
[tex]\[ 4^2 - 4 \cdot 1 \cdot (-3) = 16 + 12 = 28 \][/tex]
So the expression becomes:
[tex]\[ x = \frac{-4 \pm \sqrt{28}}{2} \][/tex]
### Step 6: Simplify the square root of 28
We can simplify [tex]\(\sqrt{28}\)[/tex]:
[tex]\[ \sqrt{28} = \sqrt{4 \cdot 7} = 2\sqrt{7} \][/tex]
### Step 7: Substitute back the simplified square root
Now our expression is:
[tex]\[ x = \frac{-4 \pm 2\sqrt{7}}{2} \][/tex]
### Step 8: Simplify the fraction
Separate the terms in the numerator to simplify:
[tex]\[ x = \frac{-4}{2} \pm \frac{2\sqrt{7}}{2} \][/tex]
[tex]\[ x = -2 \pm \sqrt{7} \][/tex]
### Step 9: Write the solutions
We have two solutions:
[tex]\[ x_1 = -2 + \sqrt{7} \][/tex]
[tex]\[ x_2 = -2 - \sqrt{7} \][/tex]
So, the solutions to the equation [tex]\(x^2 + 4x = 3\)[/tex] are:
[tex]\[ x = -2 + \sqrt{7} \][/tex]
[tex]\[ x = -2 - \sqrt{7} \][/tex]
These are the solutions for the given quadratic equation.
