How do you use the quadratic formula to solve #2x^2-5x-2=0#?

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Answer 1 · 2017-08-24T17:22:34

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

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

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

#x = (5 +- sqrt(25 - (-16)))/4#

#x = (5 +- sqrt(25 + 16))/4#

#x = (5 +- sqrt(41))/4#