What are the solutions to the equation t^2+t=-7 ?

The solutions to the equation t^2+t=-7 are t=-1/2+i(3sqrt(3))/2 ,t=-1/2-i(3sqrt(3))/2

Find the zeros of t^2+t=-7

The zeros of t^2+t=-7 are t=-1/2+i(3sqrt(3))/2 ,t=-1/2-i(3sqrt(3))/2

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t^{2} + t = - 7

t = - \frac{1}{2} + i \frac{3 3}{2}, t = - \frac{1}{2} - i \frac{3 3}{2}

Solution steps

t^{2} + t = - 7

Move

7

to the left side

t^{2} + t + 7 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot 7}{2 \cdot 1}

Simplify

1^{2} - 4 \cdot 1 \cdot 7 : 33 i

t_{1, 2} = \frac{- 1 \pm 3 3 i}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 1 + 3 3 i}{2 \cdot 1}, t_{2} = \frac{- 1 - 3 3 i}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

t = - \frac{1}{2} + i \frac{3 3}{2}, t = - \frac{1}{2} - i \frac{3 3}{2}

x^{2} - 4 x + 3 = x - 1

r^{2} - 2 r + 17 = 0

x^{2} + 3 x + 5 = 2 x + 7

x^{2} = 7 x

5 r^{2} - 12 r + 6 = 2

What are the solutions to the equation t^2+t=-7 ?
The solutions to the equation t^2+t=-7 are t=-1/2+i(3sqrt(3))/2 ,t=-1/2-i(3sqrt(3))/2
Find the zeros of t^2+t=-7
The zeros of t^2+t=-7 are t=-1/2+i(3sqrt(3))/2 ,t=-1/2-i(3sqrt(3))/2