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solve quadratic formulax^2-4x+3=0

Sagot :

The general formula for a equation of the form:

[tex]ax^2+bx+c=0[/tex]

is:

[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]

In this case we notice that a=1, b=-4 and c=3. Plugging this values in the general formula we get:

[tex]\begin{gathered} x=\frac{-(-4)\pm\sqrt[]{(-4)^2-4(1)(3)}}{2(1)} \\ =\frac{4\pm\sqrt[]{16-12}}{2} \\ =\frac{4\pm\sqrt[]{4}}{2} \\ =\frac{4\pm2}{2} \end{gathered}[/tex]

then:

[tex]x_1=\frac{4+2}{2}=\frac{6}{2}=3[/tex]

and

[tex]x_2=\frac{4-2}{2}=\frac{2}{2}=1[/tex]

Therefore, x=3 or x=1.