What are the solutions to the equation t^2+4t-25=0 ?

The solutions to the equation t^2+4t-25=0 are t=-2+sqrt(29),t=-2-sqrt(29)

Find the zeros of t^2+4t-25=0

The zeros of t^2+4t-25=0 are t=-2+sqrt(29),t=-2-sqrt(29)

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t^{2} + 4 t - 25 = 0

t = - 2 + 29, t = - 2 - 29

Decimal Notation

t = 3.38516 \dots, t = - 7.38516 \dots

Solution steps

t^{2} + 4 t - 25 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 25 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 25) = 229

t_{1, 2} = \frac{- 4 \pm 2 29}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 4 + 2 29}{2 \cdot 1}, t_{2} = \frac{- 4 - 2 29}{2 \cdot 1}

t = \frac{- 4 + 2 29}{2 \cdot 1} : - 2 + 29

t = \frac{- 4 - 2 29}{2 \cdot 1} : - 2 - 29

The solutions to the quadratic equation are:

t = - 2 + 29, t = - 2 - 29

10 x = 100

x^{9} \cdot x^{- 7}

5. 5^{2}

2 s + 5 = 0

- 3 x^{2} + 27 = 0

What are the solutions to the equation t^2+4t-25=0 ?
The solutions to the equation t^2+4t-25=0 are t=-2+sqrt(29),t=-2-sqrt(29)
Find the zeros of t^2+4t-25=0
The zeros of t^2+4t-25=0 are t=-2+sqrt(29),t=-2-sqrt(29)