What are the solutions to the equation t^2+2t-8=0 ?

The solutions to the equation t^2+2t-8=0 are t=2,t=-4

Find the zeros of t^2+2t-8=0

The zeros of t^2+2t-8=0 are t=2,t=-4

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t^{2} + 2 t - 8 = 0

t = 2, t = - 4

Solution steps

t^{2} + 2 t - 8 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 8) = 6

t_{1, 2} = \frac{- 2 \pm 6}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 2 + 6}{2 \cdot 1}, t_{2} = \frac{- 2 - 6}{2 \cdot 1}

t = \frac{- 2 + 6}{2 \cdot 1} : 2

t = \frac{- 2 - 6}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

t = 2, t = - 4

0 = - 25 t^{2} + 200 t + 375

\frac{z}{z - 5} \geq 2 z

27 x - 9

4^{2 x + 6} \leq 6 4^{2 x - 4}

\frac{2}{13 + 11}

What are the solutions to the equation t^2+2t-8=0 ?
The solutions to the equation t^2+2t-8=0 are t=2,t=-4
Find the zeros of t^2+2t-8=0
The zeros of t^2+2t-8=0 are t=2,t=-4