Answer:
The solution of the equation is at x=1 and 3.
Step-by-step explanation:
Given : Equation [tex]x^2-4x+3=0[/tex]
To find : Solve the equation using the quadratic formula?
Solution :
The general form of the quadratic equation is [tex]ax^2+bx+c=0[/tex] having solution [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
On comparing with given equation, [tex]x^2-4x+3=0[/tex]
Here, a=1 , b=-4 and c=3
Substitute in the solution,
[tex]x=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(3)}}{2(1)}[/tex]
[tex]x=\frac{4\pm\sqrt{16-12}}{2}[/tex]
[tex]x=\frac{4\pm\sqrt{4}}{2}[/tex]
[tex]x=\frac{4\pm2}{2}[/tex]
[tex]x=2\pm1[/tex]
[tex]x=3,1[/tex]
Therefore, The solution of the equation is at x=1 and 3.
