How do you solve #6x^{2}-27x-5=0#?

1 Answer
Nov 29, 2017

See a solution process below;

Explanation:

We can use the quadratic equation to solve this problem:

The quadratic formula states:

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)(-27)# for #color(blue)(b)#

#color(green)(-5)# for #color(green)(c)# gives:

#x = (-color(blue)(-27) +- sqrt(color(blue)((-27))^2 - (4 * color(red)(6) * color(green)(-5))))/(2 * color(red)(6))#

#x = (27 +- sqrt(729 - (-120)))/12#

#x = (27 +- sqrt(729 + 120))/12#

#x = (27 +- sqrt(849))/12#