How do you solve #y = 0.04x+ 8.3x + 4.3# using the quadratic formula?

1 Answer
Nov 7, 2017

See a solution process below:

Explanation:

Assuming the equation is:

#y = 0.04x^color(red)(2) + 8.3x + 4.3#

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)(0.04)# for #color(red)(a)#

#color(blue)(8.3)# for #color(blue)(b)#

#color(green)(4.3)# for #color(green)(c)# gives:

#x = (-color(blue)(8.3) +- sqrt(color(blue)(8.3)^2 - (4 * color(red)(0.04) * color(green)(4.3))))/(2 * color(red)(0.04))#

#x = (-color(blue)(8.3) +- sqrt(68.89 - 0.688))/0.08#

#x = (-8.3 +- sqrt(68.202))/0.08#

If it is necessary to get to a single number:

#x = (-8.3 - 8.258)/0.08# and #x = (-8.3 + 8.258)/0.08#

#x = -16.558/0.08# and #x = -0.042/0.08#

#x = -206.975# and #x = --0.525#

rounded to the nearest thousandth