How do you solve #x^2-10x+25=196#?

1 Answer
Apr 10, 2016

Answer:

See explanation...

Explanation:

#x^2-10x+25=196#
In order to solve quadratic equation you have to make it equal to zero.
#x^2-10x+25-196=0#
#x^2-10x-171=0#

Then calculate the discriminant:
#[DELTA]=b^2-4*a*c =(-10)^2-4*1*(-171)= =100+684= =784#
*DELTA iz a sign for discriminant.

Write a quadratic formula and input all numbers in it, and solve for x1,2.

#x1,2= (-b +- sqrt(DELTA))/(2a)#
#x1,2=(-(-10) +- sqrt(784))/(2*1)#
#x1,2=(10 +- 28)/2#
#x1=(10+28)/2=38/2=19#
#x2=(10-28)/2=-18/2=-9#

Accordingly, there are two roots since the equation is quadratic.
The roots are: #x1=19# and #x2=-9#.