How do you solve #75x ^ { 2} + 15x - 18= 0#?

1 Answer
Sep 18, 2017

See a solution process below:

Explanation:

First, we can factor a #color(red)(3)# out of each term of the equation:

#(color(red)(3) * 25x^2) + (color(red)(3) * 5x) - (color(red)(3) * 6) = 0#

#color(red)(3)(25x^2 + 5x - 6) = 0#

#(color(red)(3)(25x^2 + 5x - 6))/3 = 0/3#

#(cancel(color(red)(3))(25x^2 + 5x - 6))/color(red)(cancel(color(black)(3))) = 0#

#25x^2 + 5x - 6 = 0#

Now, 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)(25)# for #color(red)(a)#

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

#color(green)(-6)# for #color(green)(c)# gives:

#x = (-color(blue)(5) +- sqrt(color(blue)(5)^2 - (4 * color(red)(25) * color(green)(-6))))/(2 * color(red)(25))#

#x = (-color(blue)(5) +- sqrt(25 - (-600)))/50#

#x = (-color(blue)(5) +- sqrt(25 + 600))/50#

#x = (-color(blue)(5) - sqrt(625))/50# and #x = (-color(blue)(5) + sqrt(625))/50#

#x = (-color(blue)(5) - 25)/50# and #x = (-color(blue)(5) + 25)/50#

#x = -30/50# and #x = 20/50#

#x = -3/5# and #x = 2/5#