First, multiply each side of the equation by #color(red)(12)# to eliminate the fractions:
#color(red)(12)(2/3x^2 + 1/4x) = color(red)(12) xx 3#
#(color(red)(12) xx 2/3x^2) + (color(red)(12) xx 1/4x) = 36#
#8x^2 + 3x = 36#
Next, put the equation in standard quadratic form:
#8x^2 + 3x - color(red)(36) = 36 - color(red)(36)#
#8x^2 + 3x - 36 = 0#
We can nowuse 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)(8)# for #color(red)(a)#
#color(blue)(3)# for #color(blue)(b)#
#color(green)(-36)# for #color(green)(c)# gives:
#x = (-color(blue)(3) +- sqrt(color(blue)(3)^2 - (4 * color(red)(8) * color(green)(-36))))/(2 * color(red)(8))#
#x = (-color(blue)(3) +- sqrt(9 - (-1152)))/16#
#x = (-color(blue)(3) +- sqrt(9 + 1152))/16#
#x = (-color(blue)(3) +- sqrt(1161))/16#
#x = (-color(blue)(3) +- sqrt(9 * 129))/16#
#x = (-color(blue)(3) +- sqrt(9)sqrt(129))/16#
#x = (-color(blue)(3) +- 3sqrt(129))/16#
#x = (3(-1 +- sqrt(129)))/16#
Or
#x = 3/16(-1 +- sqrt(129))#
