Question

`3k^2+7k+1=0` ?

use the quadratic formula to solve the equation

Answer 1

`k = 1/6(-7+-sqrt(37))`

Explanation:

Given:

`3k^2+7k+1=0`

Multiply by `2^2*3 = 12` and complete the square as follows:

`0 = 12(3k^2+7k+1)`

`color(white)(0) = 36k^2+84k+12`

`color(white)(0) = (6k)^2+2(6k)(7)+(7)^2-37`

`color(white)(0) = (6k+7)^2-(sqrt(37))^2`

`color(white)(0) = ((6k+7)-sqrt(37))((6k+7)+sqrt(37))`

`color(white)(0) = (6k+7-sqrt(37))(6k+7+sqrt(37))`

Hence:

`6k = -7+-sqrt(37)`

So:

`k = 1/6(-7+-sqrt(37))`

Answer 2

The quadratic formula for the roots, solutions, zeros or x-intercepts of a quadratic function is as follows:
`k= (-B+-sqrt(B^2-4AC))/(2A)`

Let's identify parts of the quadratic function given in its standard form: `ax^2+bx+c`
A=3
B=7
C=1

So then substitute:
`k= (-(7)+-sqrt((7)^2-4(3)(1)))/(2*(3))`
`k= (-(7)+-sqrt(49-12))/(2*(3))`
`k= (-7+-sqrt(37))/(6)`

I'd leave it like that unless you want approximate values for K
For approximate k values plug in seperately:
`k= (-7+sqrt(37))/(6)`
`k= (-7-sqrt(37))/(6)`

There will be two solutions