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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- 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 ( - 1 ) \cdot 8}{2 ( - 1 )}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

t = - 2, t = 4

x^{2} = 3 x + 54

x, 4 x^{2} - 4 x + 4 y^{2} - k + 20 = 0

3 w^{2} + 18 w = 0

x, 16 \frac{x}{2} x = - \frac{1}{y}

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