Let's expand and simplify the left side of the equation:
(x+5)^2 = x^2 + 10x + 2(x-2)^2 = x^2 - 4x + (x-7)(x+7) = x^2 - 49
Adding them together:
x^2 + 10x + 25 + x^2 - 4x + 4 + x^2 - 49 = 11x + 80
Combining like terms:
3x^2 + 6x - 20 = 11x + 80
Now let's move all terms to one side to set the equation to zero:
3x^2 + 6x - 20 - 11x - 80 = 3x^2 - 5x - 100 = 0
This is a quadratic equation that can be solved using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values of a, b, and c:
x = (5 ± √((-5)^2 - 4(3)(-100))) / 2(3x = (5 ± √(25 + 1200)) / x = (5 ± √1225) / x = (5 ± 35) / 6
There are two possible solutions:
x = (5 + 35) / 6 = 40 / 6 = 20/3 or x = 6.6x = (5 - 35) / 6 = -30 / 6 = -5
Therefore, the solutions for the equation are x = 20/3 or x = -5.
Let's expand and simplify the left side of the equation:
(x+5)^2 = x^2 + 10x + 2
(x-2)^2 = x^2 - 4x +
(x-7)(x+7) = x^2 - 49
Adding them together:
x^2 + 10x + 25 + x^2 - 4x + 4 + x^2 - 49 = 11x + 80
Combining like terms:
3x^2 + 6x - 20 = 11x + 80
Now let's move all terms to one side to set the equation to zero:
3x^2 + 6x - 20 - 11x - 80 =
3x^2 - 5x - 100 = 0
This is a quadratic equation that can be solved using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values of a, b, and c:
x = (5 ± √((-5)^2 - 4(3)(-100))) / 2(3
x = (5 ± √(25 + 1200)) /
x = (5 ± √1225) /
x = (5 ± 35) / 6
There are two possible solutions:
x = (5 + 35) / 6 = 40 / 6 = 20/3 or x = 6.6
x = (5 - 35) / 6 = -30 / 6 = -5
Therefore, the solutions for the equation are x = 20/3 or x = -5.