What are the solutions to the equation -2t^2-4t+16=0 ?

The solutions to the equation -2t^2-4t+16=0 are t=-4,t=2

Find the zeros of -2t^2-4t+16=0

The zeros of -2t^2-4t+16=0 are t=-4,t=2

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- 2 t^{2} - 4 t + 16 = 0

t = - 4, t = 2

Solution steps

- 2 t^{2} - 4 t + 16 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 2 ) \cdot 16}{2 ( - 2 )}

(- 4)^{2} - 4 (- 2) \cdot 16 = 12

t_{1, 2} = \frac{- ( - 4 ) \pm 12}{2 ( - 2 )}

Separate the solutions

t_{1} = \frac{- ( - 4 ) + 12}{2 ( - 2 )}, t_{2} = \frac{- ( - 4 ) - 12}{2 ( - 2 )}

t = \frac{- ( - 4 ) + 12}{2 ( - 2 )} : - 4

t = \frac{- ( - 4 ) - 12}{2 ( - 2 )} : 2

The solutions to the quadratic equation are:

t = - 4, t = 2

(x + 2) (x - 1) + 26 < (x + 4) (x + 5)

108

20 x + 40 = 40 x + 20

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What are the solutions to the equation -2t^2-4t+16=0 ?
The solutions to the equation -2t^2-4t+16=0 are t=-4,t=2
Find the zeros of -2t^2-4t+16=0
The zeros of -2t^2-4t+16=0 are t=-4,t=2