What are the solutions to the equation t^2+15t=-36 ?

The solutions to the equation t^2+15t=-36 are t=-3,t=-12

Find the zeros of t^2+15t=-36

The zeros of t^2+15t=-36 are t=-3,t=-12

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t^{2} + 15 t = - 36

t = - 3, t = - 12

Solution steps

t^{2} + 15 t = - 36

Move

36

to the left side

t^{2} + 15 t + 36 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

1 5^{2} - 4 \cdot 1 \cdot 36 = 9

t_{1, 2} = \frac{- 15 \pm 9}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 15 + 9}{2 \cdot 1}, t_{2} = \frac{- 15 - 9}{2 \cdot 1}

t = \frac{- 15 + 9}{2 \cdot 1} : - 3

t = \frac{- 15 - 9}{2 \cdot 1} : - 12

The solutions to the quadratic equation are:

t = - 3, t = - 12

(- 2)^{3}

x + 10 = 5 x + 2

x^{2} = \frac{5}{6} x - \frac{1}{8}

12 x = 19

∣ 2 x - 1 ∣ = 7

What are the solutions to the equation t^2+15t=-36 ?
The solutions to the equation t^2+15t=-36 are t=-3,t=-12
Find the zeros of t^2+15t=-36
The zeros of t^2+15t=-36 are t=-3,t=-12