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

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

Find the zeros of t^2+2t=0

The zeros of t^2+2t=0 are t=0,t=-2

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t^{2} + 2 t = 0

t = 0, t = - 2

Solution steps

t^{2} + 2 t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot 0 = 2

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

Separate the solutions

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

t = \frac{- 2 + 2}{2 \cdot 1} : 0

t = \frac{- 2 - 2}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

t = 0, t = - 2

21.3 \cdot \frac{1}{137.33} \cdot \frac{1}{3} \cdot \frac{26.98}{1}

\frac{( 2 ) ^{2}}{2}

∣ x - 2 ∣ = 6

0 < x^{2} < 1

x (x - 2) (x + 8) < 0

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