How do you solve #14x ^ { 2} - 29x - 108= 0#?

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Answer 1 · 2017-08-27T14:30:12

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)(14)# for #color(red)(a)#

#color(blue)(-29)# for #color(blue)(b)#

#color(green)(-108)# for #color(green)(c)# gives:

#x = (-color(blue)((-29)) +- sqrt(color(blue)((-29))^2 - (4 * color(red)(14) * color(green)(-108))))/(2 * color(red)(14))#

#x = (color(blue)(29) +- sqrt(841 - (-6048)))/28#

#x = (color(blue)(29) +- sqrt(841 + 6048))/28#

#x = (color(blue)(29) +- sqrt(6889))/28#

#x = (color(blue)(29) +- 83)/28#

#x = (color(blue)(29) - 83)/28# and #x = (color(blue)(29) + 83)/28#

#x = -54/28# and #x = 112/28#

#x = (2 xx -27)/(2 xx 14)# and #x = 4#

#x = (color(red)(cancel(color(black)(2))) xx -27)/(color(red)(cancel(color(black)(2))) xx 14)# and #x = 4#

#x = -27/ 14# and #x = 4#