What are the solutions to the equation t^2+8t+15=0 ?

The solutions to the equation t^2+8t+15=0 are t=-3,t=-5

Find the zeros of t^2+8t+15=0

The zeros of t^2+8t+15=0 are t=-3,t=-5

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t^{2} + 8 t + 15 = 0

t = - 3, t = - 5

Solution steps

t^{2} + 8 t + 15 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot 15}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot 15 = 2

t_{1, 2} = \frac{- 8 \pm 2}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 8 + 2}{2 \cdot 1}, t_{2} = \frac{- 8 - 2}{2 \cdot 1}

t = \frac{- 8 + 2}{2 \cdot 1} : - 3

t = \frac{- 8 - 2}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

t = - 3, t = - 5

\frac{1}{2} - (\frac{5}{3} - 1) - \frac{1}{2} + \frac{5}{3}

x (5 - 6 x) \leq 0

\frac{n}{2} + 5 = 3

- 8 x - 3 = - 2 x + 1

- \frac{2}{n - 4} - \frac{n}{2 n - 8}

What are the solutions to the equation t^2+8t+15=0 ?
The solutions to the equation t^2+8t+15=0 are t=-3,t=-5
Find the zeros of t^2+8t+15=0
The zeros of t^2+8t+15=0 are t=-3,t=-5