What are the solutions to the equation t^2+22t=-27 ?

The solutions to the equation t^2+22t=-27 are t=-11+sqrt(94),t=-11-sqrt(94)

Find the zeros of t^2+22t=-27

The zeros of t^2+22t=-27 are t=-11+sqrt(94),t=-11-sqrt(94)

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t^{2} + 22 t = - 27

t = - 11 + 94, t = - 11 - 94

Decimal Notation

t = - 1.30464 \dots, t = - 20.69535 \dots

Solution steps

t^{2} + 22 t = - 27

Move

27

to the left side

t^{2} + 22 t + 27 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 22 \pm 2 2 ^{2} - 4 \cdot 1 \cdot 27}{2 \cdot 1}

2 2^{2} - 4 \cdot 1 \cdot 27 = 294

t_{1, 2} = \frac{- 22 \pm 2 94}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 22 + 2 94}{2 \cdot 1}, t_{2} = \frac{- 22 - 2 94}{2 \cdot 1}

t = \frac{- 22 + 2 94}{2 \cdot 1} : - 11 + 94

t = \frac{- 22 - 2 94}{2 \cdot 1} : - 11 - 94

The solutions to the quadratic equation are:

t = - 11 + 94, t = - 11 - 94

(3 n - 1) \cdot x^{2} - (5 n + 2) \cdot x + 2 = 0

356 = 500 - 4 q^{2}

c^{2} - 6 = - c

3 x^{2} - 12 x - 8 = 0

27 + 20 x (x - 2) = (5 x - 4)^{2} - (3 x + 5) (2 x - 1)

What are the solutions to the equation t^2+22t=-27 ?
The solutions to the equation t^2+22t=-27 are t=-11+sqrt(94),t=-11-sqrt(94)
Find the zeros of t^2+22t=-27
The zeros of t^2+22t=-27 are t=-11+sqrt(94),t=-11-sqrt(94)