A tangent line of #y = f(x)# at some point #(x_0, f(x_0))# on the curve will be a line with a slope of #f'(x_0)#. Plugging this into the point-slope formula of a line, we get
#y - f(x_0) = f'(x_0)(x - x_0)#
#=> y = f'(x_0) - x_0f'(x_0) + f(x_0)#
Given #f(x) = 5x^2#, we get #f'(x) = 10x#, meaning a tangent line at #(x, y) = (x_0, 5x_0^2)# will be a line of the form
#y = (10x_0)x - x_0(10x_0) + 5x_0^2#
#=> y = 10x_0x - 5x_0^2#
We wish to find any such lines which pass through the point #(-1, -1)#. Substituting in #(x, y) = (-1, -1)# gives us
#(-1) = 10x_0(-1) - 5x_0^2#
#=> 5x_0^2 + 10x_0 - 1 = 0#
#=> x_0 = (-5+-sqrt(10^2-4(-1)(5)))/(2(5))#
#=(-5+-2sqrt(30))/10#
#=1/2+-sqrt(30)/5#
Thus, substituting those values back into our equation for a tangent line #y = 10x_0x - 5x_0^2#, we get our two tangent lines as
#y = (5+2sqrt(30))x - 5(1/2+sqrt(30)/5)^2#
and
#y = (5-2sqrt(30))x - 5(1/2-sqrt(30)/5)^2#
