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

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Answer 1 · 2016-04-10T18:43:41

See 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 $x2=-9$ .

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