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

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

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

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

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

t = 3, t = 2

Solution steps

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

Switch sides

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

x, 5 - x^{2} + 4 y^{3} = 7 y

4 x^{2} - 12 x - 3 = 0

- 2 x^{2} + 4 = 12

x^{2} + 12 x - 28 = 0

(x + 6) (x + 2) = 5

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