We have y = 2x^2 - 452x - 68, where:
a = 2,
b = -452, and
c = -68.
Plug them in to the quadratic formula:
x = (-b pm sqrt(b^2 - 4ac))/(2a)
x = (-(-452) pm sqrt((-452)^2 - 4(2)(-68)))/(2(2))
We'll evaluate this starting from what's inside the square root. However, I think we should do some prime factorization.
452 = 2^2 * 113
(-452)^2 = 2^4 * 113^2
Substitute this back in:
x = (-(-452) pm sqrt(2^4 * 113^2 - 4(2)(-68)))/(2(2))
Then do the same for -4(2)(-68):
-4 * 2 * -68 = 4 * 2 * 68
4 * 2 * 68 = (2^2) * (2^1) * (2^2 * 17) = 2^5 * 17
So we have:
x = (-(-452) pm sqrt(2^4 * 113^2 + 2^5 * 17))/(2(2))
Factor out what's common, which is 2^4:
x = (-(-452) pm sqrt(2^4(113^2 + 2 * 17)))/(2(2))
Then evaluate sqrt(2^4):
x = (-(-452) pm 2^2 sqrt(113^2 + 2 * 17))/(2(2))
x = (-(-452) pm 4 sqrt(113^2 + 2 * 17))/(2(2))
Now, let's look at what's outside the square root. -(-452) becomes 452, and 2(2) becomes 4:
x = (452 pm 4 sqrt(113^2 + 2 * 17))/4
On the numerator, we can factor out 4:
x = (4(113 pm sqrt(113^2 + 2 * 17)))/4
And they cancel out:
x = 113 pm sqrt(113^2 + 2 * 17)
Let's evaluate what's inside the square root:
113^2 + 2 * 17 = 12769 + 2 * 17 = 12769 + 34 = 12803
x = 113 pm sqrt(12803)
So we have two real solutions:
x_1 = 113 + sqrt(12803)
x_2 = 113 - sqrt(12803)
And no imaginary solutions. We could estimate the value of the square root to get:
x_1 ~~ 113 + 113.15 = 226.15
x_2 ~~ 113 - 113.15 = -0.15