What are the solutions to the equation t^2-9t+18=0 ?

The solutions to the equation t^2-9t+18=0 are t=6,t=3

Find the zeros of t^2-9t+18=0

The zeros of t^2-9t+18=0 are t=6,t=3

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t^{2} - 9 t + 18 = 0

t = 6, t = 3

Solution steps

t^{2} - 9 t + 18 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot 18}{2 \cdot 1}

(- 9)^{2} - 4 \cdot 1 \cdot 18 = 3

t_{1, 2} = \frac{- ( - 9 ) \pm 3}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 9 ) + 3}{2 \cdot 1}, t_{2} = \frac{- ( - 9 ) - 3}{2 \cdot 1}

t = \frac{- ( - 9 ) + 3}{2 \cdot 1} : 6

t = \frac{- ( - 9 ) - 3}{2 \cdot 1} : 3

The solutions to the quadratic equation are:

t = 6, t = 3

x^{2} + 7 x + 12 = 0

- 11 + 4 x^{2} = - 7 x - 2

5 x^{2} - 2 - x^{2} = 3 x^{2} - 3

6 x^{2} = 36 x - 150

x, lo g_{10} (y) - x^{2} - y = 0

What are the solutions to the equation t^2-9t+18=0 ?
The solutions to the equation t^2-9t+18=0 are t=6,t=3
Find the zeros of t^2-9t+18=0
The zeros of t^2-9t+18=0 are t=6,t=3