Question

Solve 2x^2 + 19x - 145 = 0?

Answer 1

See a solution process below:

Explanation:

We can use the quadratic equation to solve this problem:

The quadratic formula states:

For `color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0`, the values of `x` which are the solutions to the equation are given by:

`x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))`

Substituting:

`color(red)(2)` for `color(red)(a)`

`color(blue)(19)` for `color(blue)(b)`

`color(green)(-145)` for `color(green)(c)` gives:

`x = (-color(blue)(19) +- sqrt(color(blue)(19)^2 - (4 * color(red)(2) * color(green)(-145))))/(2 * color(red)(2))`

`x = (-19 +- sqrt(361 - (8 * color(green)(-145))))/4`

`x = (-19 +- sqrt(361 - (-1160)))/4`

`x = (-19 +- sqrt(361 + 1160))/4`

`x = (-19 +- sqrt(1521))/4`

`x = (-19 - 39)/4` and `x = (-19 + 39)/4`

`x = (-58)/4` and `x = 20/4`

`x = -14.5` and `x = 5`

The Solution Set Is: `x = {-14.5, 5}`

Answer 2

See details below....

Explanation:

`2x^2+19x-145=0`

Start by factoring the the left side

`(2x + 29)(x-5)`

Then set factors equal to `0`

`2x + 29 = 0 or x-5=0`

`2x = 0 - 29 or x= 0 + 5`

`2x = -29 or x=5`

`x = (-29)/2 or x=5`

Answer 3

By using the quadratic formula, we find that x=5 and x=-14.5

Explanation:

The quadratic formula takes an equation that looks like this:

`ax^2+bx+c`

And plugs it into a formula that solves for x:

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

Based on our equation, we know the values of a, b, and c:

`a=2`
`b=19`
`c=-145`

`(-19+-sqrt(19^2-4(2xx-145)))/(2(2))`

`(-19+-sqrt(361+1160))/4 rArr (-19+-sqrt(1521))/4`

`(-19+-39)/4 rArr x=[5, -14.5]`