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#
