What are the solutions to the equation t^2-19t+18=0 ?

The solutions to the equation t^2-19t+18=0 are t=18,t=1

Find the zeros of t^2-19t+18=0

The zeros of t^2-19t+18=0 are t=18,t=1

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t^{2} - 19 t + 18 = 0

t = 18, t = 1

Solution steps

t^{2} - 19 t + 18 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 19 ) \pm ( - 19 ) ^{2} - 4 \cdot 1 \cdot 18}{2 \cdot 1}

(- 19)^{2} - 4 \cdot 1 \cdot 18 = 17

t_{1, 2} = \frac{- ( - 19 ) \pm 17}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 19 ) + 17}{2 \cdot 1}, t_{2} = \frac{- ( - 19 ) - 17}{2 \cdot 1}

t = \frac{- ( - 19 ) + 17}{2 \cdot 1} : 18

t = \frac{- ( - 19 ) - 17}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

t = 18, t = 1

4 m^{2} + 2 m = 7

(\frac{1}{2} x^{3} y^{2} - 4 y)^{9}

2 x^{4} + 7 x^{3} - 87 x^{2} - 395 x - 175

(x - 2)^{2} = 16

1 = x^{2} - 6 x + 9

What are the solutions to the equation t^2-19t+18=0 ?
The solutions to the equation t^2-19t+18=0 are t=18,t=1
Find the zeros of t^2-19t+18=0
The zeros of t^2-19t+18=0 are t=18,t=1