# real solutionsFind all real solutions to the quadratic equation  x2 + 2 = x + 5

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to find the real solutions of x^2 + 2 = x + 5

x^2 + 2 = x + 5

=> x^2 - x - 3 = 0

x1 = 1/2 + sqrt (1 + 12)/2

=> 1/2 + (sqrt 13)/2

x2 = 1/2 - (sqrt 13)/2

The roots of the equation are 1/2 + (sqrt 13)/2 and 1/2 - (sqrt 13)/2

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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The quadratic equation x^2 + 2 = x + 5 has to be solved.

x^2 + 2 = x + 5

Subtract x - 5 from both the sides

`x^2 - x - 3 = 0`

`x^2 - 2*(x/2) + 1/4 - 3 = 1/4`

`x^2 - 2*(x/2) + 1/4 = 3 + 1/4`

`x^2 - 2*(x/2) + 1/4 = 13/4`

`(x - 1/2)^2 = 13/4`

`x - 1/2 = +-sqrt 13/2`

x = `1/2 + sqrt 13/2` and x = `1/2 - sqrt 13/2`

The solution of the equation x^2 + 2 = x + 5 is `x = 1/2 + sqrt 13/2` and `x = 1/2 - sqrt 13/2`

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll re-arrange the terms of the quadratic by subtracting x + 5 both sides:

x^2 + 2 - x - 5 = 0

We'll combine like terms:

x^2 - x - 3 = 0

x1 = [-b+sqrt(b^2 - 4ac)]/2a

x2 = [-b-sqrt(b^2 - 4ac)]/2a

We'll identify a,b,c:

a = 1 , b = -1 , c = -3

We'll substitute the coefficients a,b,c into the formula of the quadratic:

x1 = [1 + sqrt(1 + 12)]/2

x1 = (1+sqrt13)/2

x2 = (1-sqrt13)/2