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

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

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

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

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t^{2} + 49 = - 14 t

t = - 7

Solution steps

t^{2} + 49 = - 14 t

Move

14 t

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

1 4^{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

9 x^{2} - 4 x + 3 = 0

8 x^{2} + 5 = 35

3 x^{(2)} + 15 x - 18 = 0

3 x^{2} - 50 x

2 x^{2} + 10 x - 25 = 0

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