Q:

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

Accepted Solution

A:
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.