Solve x^2 - 29x + 100=0?

x^2 - 29x + 100 = 0
x^2 - 4x - 25x + 100 = 0
x(x - 4) - 25(x - 4) = 0
(x - 25)(x - 4) = 0
x = 25 or x = 4

x^2 - 29x + 100 = 0
(x - 25) (x - 4) = 0
Solutions:
x = 4
x = 25

 x^2 - 29x + 100 = 0
 (x - 25) (x - 4) = 0
 Solutions:
 x = 4
 x = 25

x^2-29x+100=0
=>
x=[29+/-sqr(29^2-400)]/2
=>
x=[29+/-sqr(441)]/2
=>
x=[29+/-21]/2
=>
x=25
or
x=4

(x-25) (x-4) =0 or x=25 or x=4

x^2-29x+100=0
(x-25)(x-4)
x=4
x=25

f(x) = x^2-29x+100 = (x-25)(x-4). Then zeros of f are x = 4 and x = 25.

Using the quadratic equation formula x = 4 or x = 25

Yes, you could use the quadratic formula or complete the square to solve this, but you can save time and effort by knowing how to factor simple quadratics like this.

You want two numbers that add to a sum -29 (the coefficient on x) and multiply to a product of +100 (the constant coefficient).  To get a negative sum and a positive product, you need both numbers to be negative.

Given that, there are only a few ways to write 100 as a product of negative integers:  (-1)(-100), (-2)(-50), (-4)(-25), (-5)(-20) and (-10)(-10).  Only -4 and -25 have the required sum, so:

    x^2 - 29x + 100 = (x - 4)(x - 25)

...and the zero-product property tells you that x must be 4 or 25 for that to be zero.

x^2 - 29x + 100 = 0 ===> x^2 + (-4 - 25) * x + ((-4) * (-25)) = 0.

===> (x - 4) * (x - 25) = 0 ===> Either: x - 4 = 0, x = 4; Or: x - 25 = 0, x = 25.

SOLUTIONS: 4, 25.

The solutions are x = (29 +/- sqrt(29^2 - 4*1*100)) / 2
That’s the standard quadratic solution, it should be memorized.

x^2 - 29x + 100=0
x = 25, 4

https://www.mathpapa.com/quadratic-formula/