How do you solve #x^2-10x+8=0# using the quadratic formula?

1 Answer
Feb 29, 2016

Answer:

#x=5+-sqrt(17)#

Explanation:

#1#. Determine the #a#, #b#, and #c# values of the equation. Recall that the general quadratic equation written in standard form is: #ax^2+bx+c=0#.

#x^2-10x+8=0#

#color(red)(1)x^2# #color(orange)(-10)x# #color(blue)(+8)=0#

#a=color(red)1color(white)(XXX)b=color(orange)(-10)color(white)(XXX)c=color(blue)(8)#

#2#. Substitute the #a#, #b#, and #c# values into the quadratic formula.

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(-(color(orange)(-10))+-sqrt((color(orange)(-10))^2-4(color(red)1)(color(blue)8)))/(2(color(red)1))#

#3#. Solve for #x#.

#x=(10+-sqrt(100-32))/2#

#x=(10+-sqrt(68))/2#

#x=(10+-2sqrt(17))/2#

#x=(2(5+-sqrt(17)))/(2(1))#

#x=(color(red)cancelcolor(black)2(5+-sqrt(17)))/(color(red)cancelcolor(black)2(1))#

#color(green)(x=5+-sqrt(17))#

#:.#, #x=5+-sqrt(17)#.