What are the solutions to the equation 8t=-(16t^2+1) ?

The solutions to the equation 8t=-(16t^2+1) are t=-1/4

Find the zeros of 8t=-(16t^2+1)

The zeros of 8t=-(16t^2+1) are t=-1/4

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8 t = - (16 t^{2} + 1)

t = - \frac{1}{4}

Decimal Notation

t = - 0.25

Solution steps

8 t = - (16 t^{2} + 1)

Expand

- (16 t^{2} + 1) : - 16 t^{2} - 1

8 t = - 16 t^{2} - 1

Switch sides

- 16 t^{2} - 1 = 8 t

Move

8 t

to the left side

- 16 t^{2} - 1 - 8 t = 0

Write in the standard form

a x^{2} + b x + c = 0

- 16 t^{2} - 8 t - 1 = 0

Solve with the quadratic formula

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

(- 8)^{2} - 4 (- 16) (- 1) = 0

t_{1, 2} = \frac{- ( - 8 ) \pm 0}{2 ( - 16 )}

t = \frac{- ( - 8 )}{2 ( - 16 )}

\frac{- ( - 8 )}{2 ( - 16 )} = - \frac{1}{4}

t = - \frac{1}{4}

The solution to the quadratic equation is:

t = - \frac{1}{4}

x^{2} + x - 18 = 0

50 x^{2} - 200 x - 250 = 0

x^{2} + (\frac{3}{4})^{2} = 1

x^{2} - 2 x - 8

x (x - 1) = 1

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