What are the solutions to the equation t^2+9t+20=0 ?

The solutions to the equation t^2+9t+20=0 are t=-4,t=-5

Find the zeros of t^2+9t+20=0

The zeros of t^2+9t+20=0 are t=-4,t=-5

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t^{2} + 9 t + 20 = 0

t = - 4, t = - 5

Solution steps

t^{2} + 9 t + 20 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 1 \cdot 20}{2 \cdot 1}

9^{2} - 4 \cdot 1 \cdot 20 = 1

t_{1, 2} = \frac{- 9 \pm 1}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 9 + 1}{2 \cdot 1}, t_{2} = \frac{- 9 - 1}{2 \cdot 1}

t = \frac{- 9 + 1}{2 \cdot 1} : - 4

t = \frac{- 9 - 1}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

t = - 4, t = - 5

x (6 x + 1) \leq 2

2 y^{2} + 5 y + 2

5 (11 x + 11) > 2 (12 x + 1)

20 - 6 (x^{2} y^{2}) + 5 (2 x y)^{2}

\frac{3}{\frac{1}{3}} + \frac{4}{\frac{1}{2}}

What are the solutions to the equation t^2+9t+20=0 ?
The solutions to the equation t^2+9t+20=0 are t=-4,t=-5
Find the zeros of t^2+9t+20=0
The zeros of t^2+9t+20=0 are t=-4,t=-5