How do you solve the equation #2x^2+3x-2=0#?

1 Answer
Aug 28, 2017

Answer:

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

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

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

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

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

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

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

#x = (-color(blue)(3) - 5)/4# and #x = (-color(blue)(3) + 5)/4#

#x = -8/4# and #x = 2/4#

#x = -2# and #x = 1/2#