What are the solutions to the equation 16t^2+8t-8=-5 ?

The solutions to the equation 16t^2+8t-8=-5 are t= 1/4 ,t=-3/4

Find the zeros of 16t^2+8t-8=-5

The zeros of 16t^2+8t-8=-5 are t= 1/4 ,t=-3/4

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16 t^{2} + 8 t - 8 = - 5

t = \frac{1}{4}, t = - \frac{3}{4}

Decimal Notation

t = 0.25, t = - 0.75

Solution steps

16 t^{2} + 8 t - 8 = - 5

Move

5

to the left side

16 t^{2} + 8 t - 3 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 16 ( - 3 )}{2 \cdot 16}

8^{2} - 4 \cdot 16 (- 3) = 16

t_{1, 2} = \frac{- 8 \pm 16}{2 \cdot 16}

Separate the solutions

t_{1} = \frac{- 8 + 16}{2 \cdot 16}, t_{2} = \frac{- 8 - 16}{2 \cdot 16}

t = \frac{- 8 + 16}{2 \cdot 16} : \frac{1}{4}

t = \frac{- 8 - 16}{2 \cdot 16} : - \frac{3}{4}

The solutions to the quadratic equation are:

t = \frac{1}{4}, t = - \frac{3}{4}

- 3 v^{2} + 10 v + 5 = - 8 v^{2}

15 d^{2} - 29 d + 8 = - 4

x^{2} + 8 = - 1

x, 12 x^{2} - 3 y^{2} + 12 z^{2} + 6 x = 15

0.534 (0.5 - x)^{2} = x^{2}

What are the solutions to the equation 16t^2+8t-8=-5 ?
The solutions to the equation 16t^2+8t-8=-5 are t= 1/4 ,t=-3/4
Find the zeros of 16t^2+8t-8=-5
The zeros of 16t^2+8t-8=-5 are t= 1/4 ,t=-3/4