How do you solve `x^2-10x+16=0` using the quadratic formula?

We know the Quadratic Formula:

If `color(red)(ax^2+bx+c=0) ,`

then, `color(red)(x=(-b+-sqrt(triangle))/(2a)and triangle=b^2-4ac)`

Comparing , `color(blue)(x^2-10x+16=0`, with quad.equn ,we get

`a=1 ,b=-10, c=16`

First we find `triangle`

`triangle=(-10)^2-4(1)(16)=100-64=36`

`=>sqrt(triangle)=sqrt(36)=6`

So,

`x=(-b+-sqrt(triangle))/(2a)=(-(-10)+-6)/(2(1))=(10+-6)/2`

`i.e.x=(10+6)/2 or x=(10-6)/2`

`=>x=16/2 or x=4/2`

`=>x=8 or x=2`

Your equation is already in quadratic form, so all we have to do is substitute. The quadratic form is as follows:
`ax^2 + bx + c = 0`

Your equation looks like this:
`x^2 - 10x + 16 = 0`

What we are trying to do is find the values of `a, b` and `c`. They are located in the same spot in your equation as they are in the quadratic form.

`ax^2 + bx + c = 0`

`(1)x^2 + (-10)x + (16) = 0`
`a = 1`
`b = -10`
`c = 16`

Now plug these in to the quadratic formula:
`x = (-b \pm sqrt(b^2-4ac)) / (2a)`

`x = (-(-10) \pm sqrt(-10^2-4(1)(16)))/ (2(1))`

`x = ((color(red)+10) \pm sqrt(color(red)100-(color(blue)64)))/ (2)`

`x = (10 \pm sqrt(100-64))/ (2)`

`x = (10 \pm sqrt(36))/ (2)`

`x = (10 \pm 6)/ (2)`

Now solve by doing addition alone and subtraction alone to get your two values:

`x = (10 color(red)+ 6)/ (2) rarr x = (16)/ (2) rarr x = 8`

`x = (10 color(red)- 6)/ (2) rarr x = 4/2 rarr x = 2`

So the solution is `x = 8,2`