What are the solutions to the equation t^2-22t+9=0 ?

The solutions to the equation t^2-22t+9=0 are t=11+4sqrt(7),t=11-4sqrt(7)

Find the zeros of t^2-22t+9=0

The zeros of t^2-22t+9=0 are t=11+4sqrt(7),t=11-4sqrt(7)

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t^{2} - 22 t + 9 = 0

t = 11 + 47, t = 11 - 47

Decimal

t = 21.58300 \dots, t = 0.41699 \dots

Solution steps

t^{2} - 22 t + 9 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 1 \cdot 9}{2 \cdot 1}

(- 22)^{2} - 4 \cdot 1 \cdot 9 = 87

t_{1, 2} = \frac{- ( - 22 ) \pm 8 7}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 22 ) + 8 7}{2 \cdot 1}, t_{2} = \frac{- ( - 22 ) - 8 7}{2 \cdot 1}

t = \frac{- ( - 22 ) + 8 7}{2 \cdot 1} : 11 + 47

t = \frac{- ( - 22 ) - 8 7}{2 \cdot 1} : 11 - 47

The solutions to the quadratic equation are:

t = 11 + 47, t = 11 - 47

100 x^{2} - 10011 x + 10.011 = 0

2 (x + 6)^{2} + 16 = 34

x^{2} - x = 56

3 x^{2} + 168 = 36 x

5 (x^{2} - 7) - 1 = 9

What are the solutions to the equation t^2-22t+9=0 ?
The solutions to the equation t^2-22t+9=0 are t=11+4sqrt(7),t=11-4sqrt(7)
Find the zeros of t^2-22t+9=0
The zeros of t^2-22t+9=0 are t=11+4sqrt(7),t=11-4sqrt(7)