How do you solve #n^2=3n-18#?

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Answer 1 · 2018-03-27T12:30:39

See a solution process below:

First, transform the equation into standard quadratic form:

#n^2 - color(red)(3n) + color(blue)(18) = 3n - color(red)(3n) - 18 + color(blue)(18)#

#n^2 - 3n + 18 = 0 - 0#

#n^2 - 3n + 18 = 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)(1)# for #color(red)(a)#

#color(blue)(-3)# for #color(blue)(b)#

#color(green)(18)# for #color(green)(c)# gives:

#x = (-color(blue)(-3) +- sqrt(color(blue)((-3))^2 - (4 * color(red)(1) * color(green)(18))))/(2 * color(red)(1))#

#x = (3 +- sqrt(9 - 72))/2#

#x = (3 +- sqrt(-63))/2#