How do you solve #-21x ^ { 2} = 15x - 24#?

1 Answer
Jan 26, 2018

See a solution process below:

Explanation:

First, put the equation into standard form and reduce the coefficients where possible::

#-21x^2 - color(red)(15x) + color(blue)(24) = 15x - color(red)(15x) - 24 + color(blue)(24)#

#-21x^2 - 15x + 24 = 0 - 0#

#(-3 xx 7x^2) + (-3 xx 5x) + (-3 xx -8) = 0#

#-3(7x^2 + 5x + (-8)) = 0#

#-3(7x^2 + 5x -8) = 0#

#(-3(7x^2 + 5x - 8))/color(red)(-3) = 0/color(red)(-3)#

#(color(red)(cancel(color(black)(-3)))(7x^2 + 5x - 8))/cancel(color(red)(-3)) = 0#

#7x^2 + 5x - 8 = 0#

We can now use 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)(7)# for #color(red)(a)#

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

#color(green)(-8)# for #color(green)(c)# gives:

#x = (-color(blue)(5) +- sqrt(color(blue)(5)^2 - (4 * color(red)(7) * color(green)(-8))))/(2 * color(red)(7))#

#x = (-color(blue)(5) +- sqrt(25 - (-224)))/14#

#x = (-5 +- sqrt(25 + 224))/14#

#x = (-5 +- sqrt(249))/14#