What are the solutions to the equation t^2-8t-4=0 ?

The solutions to the equation t^2-8t-4=0 are t=2(2+sqrt(5)),t=2(2-sqrt(5))

Find the zeros of t^2-8t-4=0

The zeros of t^2-8t-4=0 are t=2(2+sqrt(5)),t=2(2-sqrt(5))

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t^{2} - 8 t - 4 = 0

t = 2 (2 + 5), t = 2 (2 - 5)

Decimal Notation

t = 8.47213 \dots, t = - 0.47213 \dots

Solution steps

t^{2} - 8 t - 4 = 0

Solve with the quadratic formula

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

(- 8)^{2} - 4 \cdot 1 \cdot (- 4) = 45

t_{1, 2} = \frac{- ( - 8 ) \pm 4 5}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 8 ) + 4 5}{2 \cdot 1}, t_{2} = \frac{- ( - 8 ) - 4 5}{2 \cdot 1}

t = \frac{- ( - 8 ) + 4 5}{2 \cdot 1} : 2 (2 + 5)

t = \frac{- ( - 8 ) - 4 5}{2 \cdot 1} : 2 (2 - 5)

The solutions to the quadratic equation are:

t = 2 (2 + 5), t = 2 (2 - 5)

x^{2} - 6 x - 81

x^{2} + 16 x + 2 = 0

9 x^{2} - 6 x + 1 = 2

s^{2} + 12 s + 20 = 0

x^{2} + 6 x + 3 = 0

What are the solutions to the equation t^2-8t-4=0 ?
The solutions to the equation t^2-8t-4=0 are t=2(2+sqrt(5)),t=2(2-sqrt(5))
Find the zeros of t^2-8t-4=0
The zeros of t^2-8t-4=0 are t=2(2+sqrt(5)),t=2(2-sqrt(5))