What are the solutions to the equation t^2-8t-5=0 ?

The solutions to the equation t^2-8t-5=0 are t=4+sqrt(21),t=4-sqrt(21)

Find the zeros of t^2-8t-5=0

The zeros of t^2-8t-5=0 are t=4+sqrt(21),t=4-sqrt(21)

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t^{2} - 8 t - 5 = 0

t = 4 + 21, t = 4 - 21

Decimal Notation

t = 8.58257 \dots, t = - 0.58257 \dots

Solution steps

t^{2} - 8 t - 5 = 0

Solve with the quadratic formula

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

(- 8)^{2} - 4 \cdot 1 \cdot (- 5) = 221

t_{1, 2} = \frac{- ( - 8 ) \pm 2 21}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 8 ) + 2 21}{2 \cdot 1}, t_{2} = \frac{- ( - 8 ) - 2 21}{2 \cdot 1}

t = \frac{- ( - 8 ) + 2 21}{2 \cdot 1} : 4 + 21

t = \frac{- ( - 8 ) - 2 21}{2 \cdot 1} : 4 - 21

The solutions to the quadratic equation are:

t = 4 + 21, t = 4 - 21

lo g_{10} (10000)

\frac{16}{2}

3 z^{2} + 5 z - 2

8 u^{3} + 125 v^{3}

4 x^{2} - 24 x + 45 = 0

What are the solutions to the equation t^2-8t-5=0 ?
The solutions to the equation t^2-8t-5=0 are t=4+sqrt(21),t=4-sqrt(21)
Find the zeros of t^2-8t-5=0
The zeros of t^2-8t-5=0 are t=4+sqrt(21),t=4-sqrt(21)