How do you solve #2x^2-3x-6=0# using the quadratic formula?

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Answer 1 · 2017-08-28T12:28:06

See a solution process below:

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

#x = (-color(blue)((-3)) +- sqrt(color(blue)((-3))^2 - (4 * color(red)(2) * color(green)(-6))))/(2 * color(red)(2))#

#x = (color(blue)(3) +- sqrt(9 - (-48)))/4#

#x = (color(blue)(3) +- sqrt(9 + 48))/4#

#x = (color(blue)(3) +- sqrt(57))/4#