To solve the quadratic equation [tex]\(3x^2 + 13x = 10\)[/tex], follow these steps:
1. Rewrite the equation in standard form:
The standard form of a quadratic equation is [tex]\(ax^2 + bx + c = 0\)[/tex].
Start with:
[tex]\[3x^2 + 13x = 10\][/tex]
Subtract 10 from both sides to move everything to one side of the equation:
[tex]\[3x^2 + 13x - 10 = 0\][/tex]
2. Identify coefficients in the standard quadratic form [tex]\(ax^2 + bx + c = 0\)[/tex]:
[tex]\[
a = 3, \quad b = 13, \quad c = -10
\][/tex]
3. Solve the quadratic equation using the quadratic formula:
The quadratic formula is:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Plug the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] into the quadratic formula:
[tex]\[
x = \frac{-13 \pm \sqrt{13^2 - 4 \cdot 3 \cdot (-10)}}{2 \cdot 3}
\][/tex]
4. Calculate the discriminant ([tex]\(b^2 - 4ac\)[/tex]):
[tex]\[
b^2 = 13^2 = 169
\][/tex]
[tex]\[
4ac = 4 \cdot 3 \cdot (-10) = -120
\][/tex]
[tex]\[
b^2 - 4ac = 169 - (-120) = 169 + 120 = 289
\][/tex]
5. Calculate the square root of the discriminant:
[tex]\[
\sqrt{289} = 17
\][/tex]
6. Plug the discriminant back into the quadratic formula:
[tex]\[
x = \frac{-13 \pm 17}{6}
\][/tex]
7. Solve for the two possible values of [tex]\(x\)[/tex]:
[tex]\[
x_1 = \frac{-13 + 17}{6} = \frac{4}{6} = \frac{2}{3}
\][/tex]
[tex]\[
x_2 = \frac{-13 - 17}{6} = \frac{-30}{6} = -5
\][/tex]
8. Conclude the solution:
The solutions to the quadratic equation [tex]\(3x^2 + 13x - 10 = 0\)[/tex] are:
[tex]\[
x = \frac{2}{3} \quad \text{and} \quad x = -5
\][/tex]
Therefore, the correct solutions are:
[tex]\[ x = \frac{2}{3} \quad \text{and} \quad x = -5 \][/tex]
So, the solution is:
[tex]\[
\boxed{x=-5 \quad \text{and} \quad x=\frac{2}{3}}
\][/tex]
