What are the solutions to the equation t^2-22t+21=0 ?

The solutions to the equation t^2-22t+21=0 are t=21,t=1

Find the zeros of t^2-22t+21=0

The zeros of t^2-22t+21=0 are t=21,t=1

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t^{2} - 22 t + 21 = 0

t = 21, t = 1

Solution steps

t^{2} - 22 t + 21 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 1 \cdot 21}{2 \cdot 1}

(- 22)^{2} - 4 \cdot 1 \cdot 21 = 20

t_{1, 2} = \frac{- ( - 22 ) \pm 20}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 22 ) + 20}{2 \cdot 1}, t_{2} = \frac{- ( - 22 ) - 20}{2 \cdot 1}

t = \frac{- ( - 22 ) + 20}{2 \cdot 1} : 21

t = \frac{- ( - 22 ) - 20}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

t = 21, t = 1

17 = r + 10

- \frac{w}{8} \leq \frac{3}{2}

\frac{b ^{2} - 81}{b + 9}

- (- 1)^{2} + 3

3 x - 17 = - 29 - 3 x

What are the solutions to the equation t^2-22t+21=0 ?
The solutions to the equation t^2-22t+21=0 are t=21,t=1
Find the zeros of t^2-22t+21=0
The zeros of t^2-22t+21=0 are t=21,t=1