Question

Solve the equation for all real solutions in simplest form.
3z^2+17z+16=0

Answer 1

`x=(-17+√(97))/(6)`,
`x=(-17-√(97))/(6)`

Explanation:

Use the quadratic formula to solve for all real solutions.

Quadratic formula:
`x=(-b\pm√(b^2-4ac))/(2a)`

Plug in and solve:

`x=(-17\pm√(17^2-4(3)(16)))/(2(3))\\\\x=(-17\pm√(289-4(3)(16)))/(2(3))\\\\x=(-17\pm√(289-12(16)))/(2(3))\\\\x=(-17\pm√(289-192))/(2(3))\\\\x=(-17\pm√(97))/(6)\\`

Since that's the farthest we can simplify to, your answers would be the positive and negative versions of the expression:

`x=(-17+√(97))/(6)`

and

`x=(-17-√(97))/(6)`