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

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

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

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

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

t = 3, t = 1

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

t = 3, t = 1

(2 - - 5)^{2}

6^{2} + 1^{2} = c^{2}

(\frac{1}{5})^{x}

(22)^{3}

5 x (x - 3) = 0

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