To solve the quadratic equation [tex]\(10x^2 + 2x - 15 = 0\)[/tex] using the quadratic formula, we'll follow these steps:
### Step 1: Identify Coefficients
First, identify the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] from the standard form of the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex].
Here, [tex]\(a = 10\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -15\)[/tex].
### Step 2: Write the Quadratic Formula
The quadratic formula is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
### Step 3: Calculate the Discriminant
The discriminant [tex]\(\Delta\)[/tex] is the part under the square root in the quadratic formula:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substitute [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] into the discriminant formula:
[tex]\[
\Delta = (2)^2 - 4(10)(-15) = 4 + 600 = 604
\][/tex]
### Step 4: Evaluate the Roots
Since the discriminant is positive ([tex]\(\Delta = 604\)[/tex]), this quadratic equation has two distinct real roots. The roots are given by:
[tex]\[
x = \frac{-b \pm \sqrt{\Delta}}{2a}
\][/tex]
Substitute the values [tex]\(\Delta = 604\)[/tex], [tex]\(a = 10\)[/tex], and [tex]\(b = 2\)[/tex] into the quadratic formula:
[tex]\[
x = \frac{-2 \pm \sqrt{604}}{20}
\][/tex]
### Step 5: Calculate the Two Solutions
#### First Solution ([tex]\(x_1\)[/tex]):
[tex]\[
x_1 = \frac{-2 + \sqrt{604}}{20}
\][/tex]
[tex]\[
x_1 \approx \frac{-2 + 24.577}{20}
\][/tex]
[tex]\[
x_1 \approx \frac{22.577}{20}
\][/tex]
[tex]\[
x_1 \approx 1.1288205727444507
\][/tex]
#### Second Solution ([tex]\(x_2\)[/tex]):
[tex]\[
x_2 = \frac{-2 - \sqrt{604}}{20}
\][/tex]
[tex]\[
x_2 \approx \frac{-2 - 24.577}{20}
\][/tex]
[tex]\[
x_2 \approx \frac{-26.577}{20}
\][/tex]
[tex]\[
x_2 \approx -1.3288205727444509
\][/tex]
### Final Solution
The solutions to the quadratic equation [tex]\(10x^2 + 2x - 15 = 0\)[/tex] are:
[tex]\[
x_1 \approx 1.1288205727444507 \quad \text{and} \quad x_2 \approx -1.3288205727444509
\][/tex]
So, summarizing, the equation:
[tex]\[
10x^2 + 2x - 15 = 0
\][/tex]
Has the solutions:
[tex]\[
x \approx 1.1288205727444507 \quad \text{or} \quad x \approx -1.3288205727444509
\][/tex]
