How do you solve #6x^2−7x−3=0 # using the quadratic formula?

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Answer 1 · 2018-05-24T12:11:30

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)(6)# for #color(red)(a)#

#color(blue)(-7)# for #color(blue)(b)#

#color(green)(-3)# for #color(green)(c)# gives:

#x = (-color(blue)(-7) +- sqrt(color(blue)(-7)^2 - (4 * color(red)(6) * color(green)(-3))))/(2 * color(red)(6))#

#x = (7 +- sqrt(49 - (-72)))/12#

#x = (7 +- sqrt(49 + 72))/12#

#x = (7 +- sqrt(121))/12#

#x = (7 +- 11)/12#

#x = (7 - 11)/12#; #x = (7 + 11)/12#

#x = (-4)/12#; #x = 18/12#

#x = -1/3#; #x = 3/2#