How do you solve #9x ^ { 2} = 4+ 7x#?

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Answer 1 · 2017-08-06T04:03:11

#x=(7+sqrt193)/18,##(7-sqrt193)/18#

Solve:

#9x^2=4+7x#

Move all terms to the left side.

#9x^2-7x-4=0# #larr# quadratic equation in standard form:

#ax^2+bx+c#, where #a=9#, #b=-7#, and #c=-4#.

Use the quadratic formula to solve.

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

Plug in the known values.

#x=(-(-7)+-sqrt((-7)^2-4*9*-4))/(2*9)#

Simplify.

#x=(7+-sqrt(49+144))/18#

#x=(7+-sqrt(193))/18#

#193# is a prime factor, so it cannot be factored furthur.

#x=(7+sqrt193)/18,##(7-sqrt193)/18#