What are the solutions to the equation 0=-16t^2+8t+12 ?

The solutions to the equation 0=-16t^2+8t+12 are t=-(-1+sqrt(13))/4 ,t=(1+sqrt(13))/4

Find the zeros of 0=-16t^2+8t+12

The zeros of 0=-16t^2+8t+12 are t=-(-1+sqrt(13))/4 ,t=(1+sqrt(13))/4

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0 = - 16 t^{2} + 8 t + 12

t = - \frac{- 1 + 13}{4}, t = \frac{1 + 13}{4}

Decimal Notation

t = - 0.65138 \dots, t = 1.15138 \dots

Solution steps

0 = - 16 t^{2} + 8 t + 12

Switch sides

- 16 t^{2} + 8 t + 12 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 16 ) \cdot 12}{2 ( - 16 )}

8^{2} - 4 (- 16) \cdot 12 = 813

t_{1, 2} = \frac{- 8 \pm 8 13}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 8 + 8 13}{2 ( - 16 )}, t_{2} = \frac{- 8 - 8 13}{2 ( - 16 )}

t = \frac{- 8 + 8 13}{2 ( - 16 )} : - \frac{- 1 + 13}{4}

t = \frac{- 8 - 8 13}{2 ( - 16 )} : \frac{1 + 13}{4}

The solutions to the quadratic equation are:

t = - \frac{- 1 + 13}{4}, t = \frac{1 + 13}{4}

(8 x + 7) (x - 6) = 0

24 x^{2} - 14 x = 0

15 x^{2} + 5 x = 0

t, x = t^{2} + 3 t

4 u^{2} = - 5 u + 6

What are the solutions to the equation 0=-16t^2+8t+12 ?
The solutions to the equation 0=-16t^2+8t+12 are t=-(-1+sqrt(13))/4 ,t=(1+sqrt(13))/4
Find the zeros of 0=-16t^2+8t+12
The zeros of 0=-16t^2+8t+12 are t=-(-1+sqrt(13))/4 ,t=(1+sqrt(13))/4