Question

How do you use the quadratic formula to solve `x^2=5-x`?

Answer 1

`x=1.79 , x=-2.79`

Explanation:

First, we need to get all the terms of the equations on one side of the equation
I added x to the left side
`x+x^2=5`
then I subtracted 5
`-5+x+x^2=0`
now I will rearrange the equation to make it look similar to `ax^2+bx+c=0`
`x^2+x-5=0`

So we want to use the quadratic formula to find what the roots of x are
the quadratic formula is
`x=(-b+-sqrt(b^2-4ac))/(2*a)`

from the equation we will identify what `a, b, and c` are
`a=1`
`b=1`
`c=-5`

now we plug all this in to the formula
`x=(-(1)+-sqrt((1)^2-4(1)(-5)))/(2*(1))`

next we simplify
`x=(-(1)+-sqrt(1+20))/(2)`

`x=(-1+-sqrt(21))/(2)`

`x=(-1+-4.58)/(2)`

there will be two answers for x because of the `+-`
`x=(-1+4.58)/(2)=3.58/2=1.79`
`x=(-1-4.58)/(2)=-5.58/2=-2.79`