Question

How do you solve `3x^2+13x-10=0`?

Answer 1

See a solution process below:

Explanation:

We can use the quadratic formula to solve this problem.

The quadratic formula states:

For `ax^2 + bx + c = 0`, the values of `x` which are the solutions to the equation are given by:

`x = (-b +- sqrt(b^2 - 4ac))/(2a)`

Substituting:

`color(red)(3)` for `color(red)(a)`

`color(blue)(13)` for `color(blue)(b)`

`color(green)(-10)` for `color(green)(c)` gives:

`x = (-color(blue)(13) +- sqrt(color(blue)(13)^2 - (4 * color(red)(3) * color(green)(-10))))/(2 * color(red)(3))`

`x = (-color(blue)(13) +- sqrt(169 - (-120)))/6`

`x = (-color(blue)(13) +- sqrt(169 + 120))/6`

`x = (-color(blue)(13) +- sqrt(289))/6`

`x = (-color(blue)(13) + 17)/6` and `x = (-color(blue)(13) - 17)/6`

`x = 4/6` and `x = -30/6`

`x = 2/3` and `x = -5`

Answer 2

`2/3` and - 5

Explanation:

`y = 3x^2 + 13x - 10 = 0`
Use the new Transforming Method (Google Search):
Transformed equation:
`y' = x^2 + 13x - 30 = 0`
Proceeding. Find 2 real roots of y', then. divide them by a = 3.
Find 2 real roots knowing the sum (-b = -13) and the product (ac = -30). They are: 2 and - 15.
Back to y, the 2 real roots are: `x1 = 2/a = 2/3` and
`x2 = -15/a = -15/3 = -5`

Answer 3

`x=2/3" "` and `" "x=-5`

Explanation:

Given:

`3x^2+13x-10=0`

Use an AC method to factor the quadratic:

Look for a pair of factors of `AC=3*10 = 30` with difference `B=13`

The pair `15,2` works.

Use this pair to split the middle term and factor by grouping:

`0 = 3x^2+13x-10`

`color(white)(0) = (3x^2+15x)-(2x+10)`

`color(white)(0) = 3x(x+5)-2(x+5)`

`color(white)(0) = (3x-2)(x+5)`

Hence zeros:

`x=2/3" "` and `" "x=-5`