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

The solutions to the equation t^2-16t+8=0 are t=2(4+sqrt(14)),t=2(4-sqrt(14))

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

The zeros of t^2-16t+8=0 are t=2(4+sqrt(14)),t=2(4-sqrt(14))

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

t = 2 (4 + 14), t = 2 (4 - 14)

Decimal Notation

t = 15.48331 \dots, t = 0.51668 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 16)^{2} - 4 \cdot 1 \cdot 8 = 414

t_{1, 2} = \frac{- ( - 16 ) \pm 4 14}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 16 ) + 4 14}{2 \cdot 1}, t_{2} = \frac{- ( - 16 ) - 4 14}{2 \cdot 1}

t = \frac{- ( - 16 ) + 4 14}{2 \cdot 1} : 2 (4 + 14)

t = \frac{- ( - 16 ) - 4 14}{2 \cdot 1} : 2 (4 - 14)

The solutions to the quadratic equation are:

t = 2 (4 + 14), t = 2 (4 - 14)

\frac{2 n - 3}{n + 1} \cdot \frac{2 n ^{2} + 5 n + 3}{9 - 4 n ^{2}}

b^{2} + 12 b + 27

\frac{v}{7} = 8

x^{2} - 17 x - 168

- 2 x + 4 < 10

What are the solutions to the equation t^2-16t+8=0 ?
The solutions to the equation t^2-16t+8=0 are t=2(4+sqrt(14)),t=2(4-sqrt(14))
Find the zeros of t^2-16t+8=0
The zeros of t^2-16t+8=0 are t=2(4+sqrt(14)),t=2(4-sqrt(14))