What are the solutions to the equation x^2-4x+3=0 ?

The solutions to the equation x^2-4x+3=0 are x=3,x=1

Find the zeros of x^2-4x+3=0

The zeros of x^2-4x+3=0 are x=3,x=1

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x^{2} - 4 x + 3 = 0

x = 3, x = 1

Solution steps

x^{2} - 4 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot 3 = 2

x_{1, 2} = \frac{- ( - 4 ) \pm 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 2}{2 \cdot 1}

x = \frac{- ( - 4 ) + 2}{2 \cdot 1} : 3

x = \frac{- ( - 4 ) - 2}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

x = 3, x = 1

(x - 2)^{2} = 9

lo g_{3} (8) = 2

(\frac{i}{3 - i}) (\frac{1}{2 + 3 i})

x^{2} - 14 x + 35 = 0

4 (x + 6) = x + 3 (x + 5) + 9

What are the solutions to the equation x^2-4x+3=0 ?
The solutions to the equation x^2-4x+3=0 are x=3,x=1
Find the zeros of x^2-4x+3=0
The zeros of x^2-4x+3=0 are x=3,x=1