What are the solutions to the equation t^2+20t=0 ?

The solutions to the equation t^2+20t=0 are t=0,t=-20

Find the zeros of t^2+20t=0

The zeros of t^2+20t=0 are t=0,t=-20

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t^{2} + 20 t = 0

t = 0, t = - 20

Solution steps

t^{2} + 20 t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

2 0^{2} - 4 \cdot 1 \cdot 0 = 20

t_{1, 2} = \frac{- 20 \pm 20}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 20 + 20}{2 \cdot 1}, t_{2} = \frac{- 20 - 20}{2 \cdot 1}

t = \frac{- 20 + 20}{2 \cdot 1} : 0

t = \frac{- 20 - 20}{2 \cdot 1} : - 20

The solutions to the quadratic equation are:

t = 0, t = - 20

p^{2} - 3 p + 5 = 0

a, (a + b + c) (a + b - c) = 3 ab

(x + 1) (x + 6) = 0

y^{2} + 24 y

x^{2} + 48 x = 0

What are the solutions to the equation t^2+20t=0 ?
The solutions to the equation t^2+20t=0 are t=0,t=-20
Find the zeros of t^2+20t=0
The zeros of t^2+20t=0 are t=0,t=-20