What are the solutions to the equation 13t^2-8t+1=-3t ?

The solutions to the equation 13t^2-8t+1=-3t are t= 5/26+i(3sqrt(3))/(26),t= 5/26-i(3sqrt(3))/(26)

Find the zeros of 13t^2-8t+1=-3t

The zeros of 13t^2-8t+1=-3t are t= 5/26+i(3sqrt(3))/(26),t= 5/26-i(3sqrt(3))/(26)

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13 t^{2} - 8 t + 1 = - 3 t

t = \frac{5}{26} + i \frac{3 3}{26}, t = \frac{5}{26} - i \frac{3 3}{26}

Solution steps

13 t^{2} - 8 t + 1 = - 3 t

Move

3 t

to the left side

13 t^{2} - 5 t + 1 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 13 \cdot 1}{2 \cdot 13}

Simplify

(- 5)^{2} - 4 \cdot 13 \cdot 1 : 33 i

t_{1, 2} = \frac{- ( - 5 ) \pm 3 3 i}{2 \cdot 13}

Separate the solutions

t_{1} = \frac{- ( - 5 ) + 3 3 i}{2 \cdot 13}, t_{2} = \frac{- ( - 5 ) - 3 3 i}{2 \cdot 13}

t = \frac{- ( - 5 ) + 3 3 i}{2 \cdot 13} : \frac{5}{26} + i \frac{3 3}{26}

t = \frac{- ( - 5 ) - 3 3 i}{2 \cdot 13} : \frac{5}{26} - i \frac{3 3}{26}

The solutions to the quadratic equation are:

t = \frac{5}{26} + i \frac{3 3}{26}, t = \frac{5}{26} - i \frac{3 3}{26}

3 x^{2} + 7 x - 49 = 0

x^{2} + x = 342

x, (x - y)^{2} + x y = 1

x, x^{2} + x y - y^{2} = 16

b^{2} + 5 b = 32 + b

What are the solutions to the equation 13t^2-8t+1=-3t ?
The solutions to the equation 13t^2-8t+1=-3t are t= 5/26+i(3sqrt(3))/(26),t= 5/26-i(3sqrt(3))/(26)
Find the zeros of 13t^2-8t+1=-3t
The zeros of 13t^2-8t+1=-3t are t= 5/26+i(3sqrt(3))/(26),t= 5/26-i(3sqrt(3))/(26)