How do you solve #6n ^ { 2} - 5n = 14#?

Question

874 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-18T13:57:04

See a solution process below:

First, subtract #color(red)(14)# from each side of the equation to put the equation in standard form:

#6n^2 - 5n - color(red)(14) = 14 - color(red)(14)#

#6n^2 - 5n - 14 = 0#

We can now use the quadratic equation to solve this problem:

For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))#

Substituting:

#color(red)(6)# for #color(red)(a)#

#color(blue)(-5)# for #color(blue)(b)#

#color(green)(-14)# for #color(green)(c)# gives:

#x = (-color(blue)(-5) +- sqrt(color(blue)(-5)^2 - (4 * color(red)(6) * color(green)(-14))))/(2 * color(red)(6))#

#x = (color(blue)(5) +- sqrt(25 - (-336)))/12#

#x = (color(blue)(5) +- sqrt(25 + 336))/12#

#x = (color(blue)(5) +- sqrt(361))/12#

#x = (color(blue)(5) +- 19)/12#

#x = (color(blue)(5) - 19)/12# and #x = (color(blue)(5) + 19)/12#

#x = -14/12# and #x = 24/12#

#x = -7/6# and #x = 2#