Question

What are the zeros `-2x^2-15x+y+22=0`?

Answer 1

`x=(-15+sqrt401)/4`, `(-15-sqrt401)/4`

Explanation:

Given:

`-2x^2-15x+y+22=0`

Subtract `y` from both sides.

`-2x^2-15x+22=-y`

Multiply both sides by `-1`. This will reverse the signs.

`2x^2+15x-22=y`

Switch sides.

`y=2x^2+15x-22`

This is a quadratic equation in standard form:

`y=ax^2+bx+c`,

where:

`a=2`, `b=15`, `c=-22`

The roots are the x-intercepts, which are the values for `x` when `y=0`.

Substitute `0` for `y`.

`0=2x^2+15x-22`

Solve for `x` using the quadratic formula:

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

Plug the known values into the equation.

`x=(-15+-sqrt(15^2-4*2*-22))/(2*2)`

`x=(-15+-sqrt(401))/4` `larr` `401` is a prime number

Roots

`x=(-15+sqrt401)/4`, `(-15-sqrt401)/4`

Approximate roots

`x~~2.56,` `-8.756`

graph{y=2x^2+15x-22 [-11.09, 11.41, -8.775, 2.475]}