What are the solutions to the equation t^2-4t-5=0 ?

The solutions to the equation t^2-4t-5=0 are t=5,t=-1

Find the zeros of t^2-4t-5=0

The zeros of t^2-4t-5=0 are t=5,t=-1

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t^{2} - 4 t - 5 = 0

t = 5, t = - 1

Solution steps

t^{2} - 4 t - 5 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 5) = 6

t_{1, 2} = \frac{- ( - 4 ) \pm 6}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 4 ) + 6}{2 \cdot 1}, t_{2} = \frac{- ( - 4 ) - 6}{2 \cdot 1}

t = \frac{- ( - 4 ) + 6}{2 \cdot 1} : 5

t = \frac{- ( - 4 ) - 6}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

t = 5, t = - 1

21 - 3 x^{2} = 0

2 x^{2} - 8 x - 10 = x + 8

4 x^{2} - 16 x + 17

y, 5 x^{2} - 2 x y + 2 y^{2} - 2 y - 71 = 0

- 2 x^{2} + 6 x - 1 = 0

What are the solutions to the equation t^2-4t-5=0 ?
The solutions to the equation t^2-4t-5=0 are t=5,t=-1
Find the zeros of t^2-4t-5=0
The zeros of t^2-4t-5=0 are t=5,t=-1