Question

How do you solve `x^2+5x-6=0` using the quadratic formula?

Answer 1

`x_1=1` or `x_2=-6`

Explanation:

Using the formula

`x_(1,2)=-p/2pmsqrt(p^2/4-q)`
`p=5,q=-6`

we get

`x_(1,2)=-5/2pmsqrt(25/4+6)`

`25/4+6=25/4+24/4=49/4`

so

`x_(1,2)-5/2pmsqrt(49/4)`
so we get

`x_1=1`

`x_2=-6`

Answer 2

`x=-6, 1`

Explanation:

Given quadratic equation: `x^2+5x-6=0`

Comparing the above equation with the standard form of quadratic equation: `ax^2+bx+c=0` we get

`a=1, b=5` & `c=-6`

The roots of quadratic equation are given as follows

`x_{1, 2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}`

`=\frac{-5\pm\sqrt{5^2-4(1)(-6)}}{2(1)}`

`=\frac{-5\pm7}{2}`

`=-6, 1`

Answer 3

`x=1,-6`

Explanation:

`x^2+5x-6=0` is a quadratic equation in standard form:

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

where:

`a=1`, `b=5`, `c=-6`

Quadratic formula

Substitute `0` for `y` and solve for `x`.

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

Plug in the known values.

`x=(-5+-sqrt(49))/2`

Simplify.

`x=(-5+-7)/2`

`x=(-5+7)/2`, `(-5-7)/2`

Simplify.

`x=2/2`, `x=-12/2`

Simplify.

`x=1,-6`