Question

Solve for x in the equation 2x^2+3x-7=x^2+5x+39

Answer 1

Subtract x^2 from both sides
x^2 + 3x - 7 = 5x + 39
Subtract 5x from both sides
x^2 - 2x - 7 = 39
Add 7 to both sides
x^2 - 2x = 46
Complete the square by adding (b/2)^2 to both sides, b = ( -2)
(-2/2) = -1, then square that (-1)^2 = 1
x^2 - 2x + 1 = 46 + 1
Simplify the expression by factoring
(x - 1)^2 = 47
Take square root on each side
x - 1 = (sqrt (47))
Solve for x
x = 1 + (sqrt (47))
Since 47 is prime, 47 cannot be broken down by the square root and this is the answer to your problem.

Answer 2

`x=1\pm√(47)`

Explanation:

We have been given an equation
`2x^2+3x-7=x^2+5x+39`. We are asked to find the solution for our given equation.

`2x^2+3x-7=x^2+5x+39`

`2x^2-x^2+3x-7=x^2-x^2+5x+39`

`x^2+3x-7=5x+39`

`x^2+3x-5x-7-39=5x-5x+39-39`

`x^2-2x-46=0`

Using quadratic formula, we will get:

`x=(-b\pm√(b^2-4ac))/(2a)`

`x=(-(-2)\pm√((-2)^2-4(1)(-46)))/(2(1))`

`x=(2\pm√(4+184))/(2)`

`x=(2\pm√(188))/(2)`

`x=(2\pm2√(47))/(2)`

`x=1\pm√(47)`

Therefore, the solutions for our given equation are
`x=1\pm√(47)`.