What are the solutions to the equation t^2-2t-3=0 ?

The solutions to the equation t^2-2t-3=0 are t=3,t=-1

Find the zeros of t^2-2t-3=0

The zeros of t^2-2t-3=0 are t=3,t=-1

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t^{2} - 2 t - 3 = 0

t = 3, t = - 1

Solution steps

t^{2} - 2 t - 3 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 3) = 4

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

Separate the solutions

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

t = \frac{- ( - 2 ) + 4}{2 \cdot 1} : 3

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

The solutions to the quadratic equation are:

t = 3, t = - 1

2 (x + 3) = 12

x^{2} + 16 > 10 x

(x - 6) (x + 1)^{2} < 0

(- x)^{2}

What are the solutions to the equation t^2-2t-3=0 ?
The solutions to the equation t^2-2t-3=0 are t=3,t=-1
Find the zeros of t^2-2t-3=0
The zeros of t^2-2t-3=0 are t=3,t=-1