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

1 Answer
Aug 17, 2017

Answer:

#x=4+sqrt(11) or x=4-sqrt(11)#

Explanation:

Given
#color(white)("XXX")x^2-8x=-10#
Re-write in standard form:
#color(white)("XXX")color(red)1x^2color(blue)(-8)x+color(green)(10)=0#

The quadratic formula tells us that an equation of the form:
#color(white)("XXX")color(red)ax^2+color(blue)bx+color(green)c=0#
has solutions:
#color(white)("XXX")x=(-color(blue)b+-sqrt(color(blue)b-4color(red)acolor(green)c))/(2color(red)a#

So in this case the solutions are
#color(white)("XXX")x=(-(color(blue)(-8))+-sqrt((color(blue)(-8))^2-4xxcolor(red)1xxcolor(green)(10)))/(2xxcolor(red)1)#

#color(white)("XXXx")=(8+-sqrt(44))/2#

#color(white)("XXXx")=(8+-2sqrt(11))/2#

#color(white)("XXXx")=4+-sqrt(11)#