What are the solutions to the equation t^2-5t+6=0 ?

The solutions to the equation t^2-5t+6=0 are t=3,t=2

Find the zeros of t^2-5t+6=0

The zeros of t^2-5t+6=0 are t=3,t=2

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t^{2} - 5 t + 6 = 0

t = 3, t = 2

Solution steps

t^{2} - 5 t + 6 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot 6 = 1

t_{1, 2} = \frac{- ( - 5 ) \pm 1}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 5 ) + 1}{2 \cdot 1}, t_{2} = \frac{- ( - 5 ) - 1}{2 \cdot 1}

t = \frac{- ( - 5 ) + 1}{2 \cdot 1} : 3

t = \frac{- ( - 5 ) - 1}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

t = 3, t = 2

1 - \frac{1}{5}

\frac{x - 1}{2} = 5

1 2^{7}

x^{4} - 6 x^{3} + 8 x^{2} + 6 x - 9

4 x + 7 \leq - 1 or 2 x - 5 > 3

What are the solutions to the equation t^2-5t+6=0 ?
The solutions to the equation t^2-5t+6=0 are t=3,t=2
Find the zeros of t^2-5t+6=0
The zeros of t^2-5t+6=0 are t=3,t=2