How do you solve #x^2+10x-9=0#?
1 Answer
Dec 13, 2016
Explanation:
#x^2+10x-9=0#
is in the form:
#ax^2+bx+c=0#
where
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)#