What are the approximate solutions of #5x^2 − 7x = 1# rounded to the nearest hundredth?

Subtracting #1# from both sides we get:

#5x^2-7x-1 = 0#

This is of the form #ax^2+bx+c = 0#, with #a = 5#, #b = -7# and #c = -1#.

The general formula for roots of such a quadratic gives us:

#x = (-b +- sqrt(b^2-4ac)) / (2a) #

#= (7+-sqrt((-7)^2-(4xx5xx-1)))/(2xx5)#

#= (7+-sqrt(69))/10#

#=0.7 +- sqrt(69)/10#

What is a good approximation for #sqrt(69)#?

We could punch it into a calculator, but let's do it by hand instead using Newton-Raphson:

#8^2 = 64#, so #8# seems like a good first approximation.

Then iterate using the formula:

#a_(n+1) = (a_n^2+69)/(2a_n)#

Let #a_0=8#

#a_1 = (64+69)/16 = 133/16 = 8.3125#

This is almost certainly good enough for the accuracy requested.

So #sqrt(69)/10 ~= 8.3 / 10 = 0.83#

#x ~= 0.7 +- 0.83#

That is #x ~= 1.53# or #x ~= -0.13#

Rewrite #5x^2-7x=1# in the standard form of #ax^2+bx+c = 0#
giving
#5x^2-7x-1 = 0#
then use the Quadratic Formula for roots:
#x = (-b+-sqrt(b^2-4ac))/(2a)#

In this case
#x = (7+-sqrt(49+20))/10#

Using a calculator:
#sqrt(69) = 8.306624# (approx.)

So
#x= 15.306624/10 = 1.53# (rounded to nearest hundredth)
or
#x = -1.306624/10 = -0.13# (rounded to the nearest hundredth)