What are the solutions to the equation 64-32t-(32/2)t^2=0 ?

The solutions to the equation 64-32t-(32/2)t^2=0 are t=-1-sqrt(5),t=sqrt(5)-1

Find the zeros of 64-32t-(32/2)t^2=0

The zeros of 64-32t-(32/2)t^2=0 are t=-1-sqrt(5),t=sqrt(5)-1

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64 - 32 t - (\frac{32}{2}) t^{2} = 0

t = - 1 - 5, t = 5 - 1

Decimal Notation

t = - 3.23606 \dots, t = 1.23606 \dots

Solution steps

64 - 32 t - (\frac{32}{2}) t^{2} = 0

Refine

64 - 32 t - 16 t^{2} = 0

Write in the standard form

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

- 16 t^{2} - 32 t + 64 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 ( - 16 ) \cdot 64}{2 ( - 16 )}

(- 32)^{2} - 4 (- 16) \cdot 64 = 325

t_{1, 2} = \frac{- ( - 32 ) \pm 32 5}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- ( - 32 ) + 32 5}{2 ( - 16 )}, t_{2} = \frac{- ( - 32 ) - 32 5}{2 ( - 16 )}

t = \frac{- ( - 32 ) + 32 5}{2 ( - 16 )} : - 1 - 5

t = \frac{- ( - 32 ) - 32 5}{2 ( - 16 )} : 5 - 1

The solutions to the quadratic equation are:

t = - 1 - 5, t = 5 - 1

15 x^{2} - x - 6

32 x^{2} - 50 y^{2}

x^{36}

\frac{a}{- 1} - 1 > 1

6 < ∣ 1 - 3 x ∣

What are the solutions to the equation 64-32t-(32/2)t^2=0 ?
The solutions to the equation 64-32t-(32/2)t^2=0 are t=-1-sqrt(5),t=sqrt(5)-1
Find the zeros of 64-32t-(32/2)t^2=0
The zeros of 64-32t-(32/2)t^2=0 are t=-1-sqrt(5),t=sqrt(5)-1