What are the solutions to the equation t^2-14t+49=0 ?

The solutions to the equation t^2-14t+49=0 are t=7

Find the zeros of t^2-14t+49=0

The zeros of t^2-14t+49=0 are t=7

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t^{2} - 14 t + 49 = 0

t = 7

Solution steps

t^{2} - 14 t + 49 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 1 \cdot 49}{2 \cdot 1}

(- 14)^{2} - 4 \cdot 1 \cdot 49 = 0

t_{1, 2} = \frac{- ( - 14 ) \pm 0}{2 \cdot 1}

t = \frac{- ( - 14 )}{2 \cdot 1}

\frac{- ( - 14 )}{2 \cdot 1} = 7

t = 7

The solution to the quadratic equation is:

t = 7

4 x^{2} - 5 = 31

n^{2} - 18 n = - 73

x^{2} - 10 x - 13 = 0

- 3 x^{2} + 11 x + 9 = 1

5.56 \cdot 1 0^{- 10} = \frac{x ^{2}}{2.5}

What are the solutions to the equation t^2-14t+49=0 ?
The solutions to the equation t^2-14t+49=0 are t=7
Find the zeros of t^2-14t+49=0
The zeros of t^2-14t+49=0 are t=7