How do you solve #x^2+10x-9=0#?

1 Answer
Dec 13, 2016

#x=-5+-sqrt(34)#

Explanation:

#x^2+10x-9=0#

is in the form:

#ax^2+bx+c=0#

where #a=1#, #b=10# and #c=-9#

Use the quadratic formula to find:

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

#color(white)(x) = (-10+-sqrt(10^2-4(1)(-9)))/(2*1)#

#color(white)(x) = (-10+-sqrt(100+36))/2#

#color(white)(x) = (-10+-sqrt(136))/2#

#color(white)(x) = (-10+-sqrt(2^2*34))/2#

#color(white)(x) = (-10+-2sqrt(34))/2#

#color(white)(x) = -5+-sqrt(34)#