What are the solutions to the equation 3t^2-14t+22=0 ?

The solutions to the equation 3t^2-14t+22=0 are t= 7/3+i(sqrt(17))/3 ,t= 7/3-i(sqrt(17))/3

Find the zeros of 3t^2-14t+22=0

The zeros of 3t^2-14t+22=0 are t= 7/3+i(sqrt(17))/3 ,t= 7/3-i(sqrt(17))/3

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3 t^{2} - 14 t + 22 = 0

t = \frac{7}{3} + i \frac{17}{3}, t = \frac{7}{3} - i \frac{17}{3}

Solution steps

3 t^{2} - 14 t + 22 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 3 \cdot 22}{2 \cdot 3}

Simplify

(- 14)^{2} - 4 \cdot 3 \cdot 22 : 217 i

t_{1, 2} = \frac{- ( - 14 ) \pm 2 17 i}{2 \cdot 3}

Separate the solutions

t_{1} = \frac{- ( - 14 ) + 2 17 i}{2 \cdot 3}, t_{2} = \frac{- ( - 14 ) - 2 17 i}{2 \cdot 3}

t = \frac{- ( - 14 ) + 2 17 i}{2 \cdot 3} : \frac{7}{3} + i \frac{17}{3}

t = \frac{- ( - 14 ) - 2 17 i}{2 \cdot 3} : \frac{7}{3} - i \frac{17}{3}

The solutions to the quadratic equation are:

t = \frac{7}{3} + i \frac{17}{3}, t = \frac{7}{3} - i \frac{17}{3}

lo g_{7} (2 - x) \leq lo g_{7} (3 x + 6)

x - 1 < \frac{2}{∣ x ∣}

7 - 2 x^{2} = - 7

(8 x^{6})^{- \frac{2}{3}}

4 (u + 2) - u = 2 (u - 1) + 9

What are the solutions to the equation 3t^2-14t+22=0 ?
The solutions to the equation 3t^2-14t+22=0 are t= 7/3+i(sqrt(17))/3 ,t= 7/3-i(sqrt(17))/3
Find the zeros of 3t^2-14t+22=0
The zeros of 3t^2-14t+22=0 are t= 7/3+i(sqrt(17))/3 ,t= 7/3-i(sqrt(17))/3