Algebra step-by-step, please?

x^2 - 5x + 3 = 0
Using quadratic formula,
x = [5±√(5^2-4*3)]/2 = [5±√(13)]/2

x^2 - 5x + 3 = 0
x = [-b ±√(b^2 - 4ac)]/(2a) (← quadratic formula)

a = 1
b = -5
c = 3

x = [5 ±√(25 - 12)]/2
x = [5 ±√13]/2

∴ x = [5 ±√13]/2

The easiest method of factoring is to use the quadratic formula

x =( -b +/- sqrt(b^2 - 4ac)) / 2a; plugging in terms of the equation in the form of ax^2 + bx + c = 0

x = (-(-5)) +/- sqrt((-5)^2 - 4(1 * 3)))/(2 * 1) =( 5 +/- sqrt(25 - 12)) / 2

This gives us the roots x = 4.3028 and x = 0.6972. Or

x - 4.3028 = 0 and x - 0.6972 = 0. So the factored equation is as follows:

(x - 0.6972)(x - 4.3028) = 0

You can multiply these terms to check that you obtain the original equation.

Question Number 1 :
For this equation x^2 - 5*x + 3 = 0 , answer the following questions :
A. Find the roots using Quadratic Formula !
B. Use completing the square to find the root of the equation !

Answer Number 1 :
The equation x^2 - 5*x + 3 = 0 is already in a*x^2+b*x+c=0 form.
So we can imply that the value of a = 1, b = -5, c = 3.

1A. Find the roots using Quadratic Formula !
Use abc formula and you get either
x1 = (-b+sqrt(b^2-4*a*c))/(2*a) or x2 = (-b-sqrt(b^2-4*a*c))/(2*a)
As we know that a = 1, b = -5 and c = 3,
we need to subtitute a,b,c in the abc formula, with thos values.
So we get x1 = (-(-5) + sqrt( (-5)^2 - 4 * (1)*(3)))/(2*1) and x2 = (-(-5) - sqrt( (-5)^2 - 4 * (1)*(3)))/(2*1)
Which can be turned into x1 = ( 5 + sqrt( 25-12))/(2) and x2 = ( 5 - sqrt( 25-12))/(2)
Which make x1 = ( 5 + sqrt( 13))/(2) and x2 = ( 5 - sqrt( 13))/(2)
So we get x1 = ( 5 + 3.60555127546399 )/(2) and x2 = ( 5 - 3.60555127546399 )/(2)
So we got the answers as x1 = 4.30277563773199 and x2 = 0.697224362268005

1B. Use completing the square to find the root of the equation !
x^2 - 5*x + 3 = 0 ,divide both side with 1
So we get x^2 - 5*x + 3 = 0 ,
The coefficient of x is -5
We have to use the fact that ( x + q )^2 = x^2 + 2*q*x + q^2 , and assume that q = -5/2 = -2.5
By using that fact we turn the equation into x^2 - 5*x + 6.25 - 3.25 = 0
And it is the same with ( x - 2.5 )^2 - 3.25 = 0
Which is the same with (( x - 2.5 ) - 1.80277563773199 ) * (( x - 2.5 ) + 1.80277563773199 ) = 0
And it is the same with ( x - 2.5 - 1.80277563773199 ) * ( x - 2.5 + 1.80277563773199 ) = 0
Do the addition/subtraction, and we get ( x - 4.30277563773199 ) * ( x - 0.697224362268005 ) = 0
We get following answers x1 = 4.30277563773199 and x2 = 0.697224362268005

x^2 - 5x + 3 = 0
{x^2- 2 5/2 x + (5/2)^2} - (5/2)^2 +3=0 Completing square
{ x - (5/2)}^2 - 25/4 +3=0
{ x - (5/2)}^2 = 25/4 -3 = 13/4
x- 5/2= +_ sqrt(13/4)
x= 5/2 +_ sqrt(13/4)
x= (5+_ sqrt 13)/2
Either x= (5+ sqrt 13)/2 OR x= (5- sqrt 13)/2x

The explanations are good, but if you need a visual example, go to www.youtube.com and search "solving a quadratic equation by the quadratic formula." The ones by minkusbc are a little dull, but informative.

x's=[5+&-(25-12)^1/2]/2=4.3 & 0.6972

No problem!

This problem is written in A^2+Bx+C form.

First, identify the coefficients.

A=1 B=5 C=3

Since this equation is not factorable, you will have to use the formula

(-B) +/- √B^2-4(A)(C)
(divided by)
2A

Or: Negative B plus the square root of B squared minus 4 times A times C. Divide the solution by @ times A.
Then: Negative B minus the square root of B squared minus 4 times A times C. Divide the solution by @ times A.
Do the math, et voila, your two solutions.

USE THE PATHAGOREAN THEOREM (spelling?)