What are the solutions to the equation 2t^2+2t+11=0 ?

The solutions to the equation 2t^2+2t+11=0 are t=-1/2+i(sqrt(21))/2 ,t=-1/2-i(sqrt(21))/2

Find the zeros of 2t^2+2t+11=0

The zeros of 2t^2+2t+11=0 are t=-1/2+i(sqrt(21))/2 ,t=-1/2-i(sqrt(21))/2

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2 t^{2} + 2 t + 11 = 0

t = - \frac{1}{2} + i \frac{21}{2}, t = - \frac{1}{2} - i \frac{21}{2}

Solution steps

2 t^{2} + 2 t + 11 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 2 \cdot 11}{2 \cdot 2}

Simplify

2^{2} - 4 \cdot 2 \cdot 11 : 221 i

t_{1, 2} = \frac{- 2 \pm 2 21 i}{2 \cdot 2}

Separate the solutions

t_{1} = \frac{- 2 + 2 21 i}{2 \cdot 2}, t_{2} = \frac{- 2 - 2 21 i}{2 \cdot 2}

t = \frac{- 2 + 2 21 i}{2 \cdot 2} : - \frac{1}{2} + i \frac{21}{2}

t = \frac{- 2 - 2 21 i}{2 \cdot 2} : - \frac{1}{2} - i \frac{21}{2}

The solutions to the quadratic equation are:

t = - \frac{1}{2} + i \frac{21}{2}, t = - \frac{1}{2} - i \frac{21}{2}

∣ 2 a + 5 ∣ = ∣ 3 a - 9 ∣

- 8 = 32 - 5 q

\frac{a}{12} = \frac{2}{3}

5 c^{2} - 6 c - 8 = 0

\frac{4 n + 7 - 5 n}{2} \geq 11

What are the solutions to the equation 2t^2+2t+11=0 ?
The solutions to the equation 2t^2+2t+11=0 are t=-1/2+i(sqrt(21))/2 ,t=-1/2-i(sqrt(21))/2
Find the zeros of 2t^2+2t+11=0
The zeros of 2t^2+2t+11=0 are t=-1/2+i(sqrt(21))/2 ,t=-1/2-i(sqrt(21))/2