Answer:
x(x-2) = 5
=> x^2 -2x-5 = 0
a=1, b= -2, c= -5
Discriminant= b^2-4ac
= (-2)^2 - 4(1)(-5)
=4+20
=24
Since Discriminant > 0
Therefore the roots are real and distinct.
Hope it helps....... mark me as brainliest
[tex]x(x - 2) = 5 \\ {x}^{2} - 2x = 5 \\ {x}^{2} - 2x + 1 = 5 + 1 \\ {(x - 1)}^{2} = 6 \\ x - 1 = \sqrt{6} \\ x = 1 + \sqrt{6} [/tex]
... Hope this will help...
