How do you solve #2p^2 - 3p -2 = 0#?

Question

3542 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-07T09:09:19

See a solution process below:

We can 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)(2)# for #color(red)(a)#

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

#color(green)(-2)# for #color(green)(c)# gives:

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

#p = (color(blue)(3) +- sqrt(9 - (-16)))/4#

#p = (color(blue)(3) +- sqrt(9 + 16))/4#

#p = (color(blue)(3) +- sqrt(25))/4#

#p = (color(blue)(3) - 5)/4#; #p = (color(blue)(3) + 5)/4#

#p = -2/4#; #p = 8/4#

#p = -1/2#; #p = 2#