What are the solutions to the equation 3t^2-16t+25=0 ?

The solutions to the equation 3t^2-16t+25=0 are t= 8/3+i(sqrt(11))/3 ,t= 8/3-i(sqrt(11))/3

Find the zeros of 3t^2-16t+25=0

The zeros of 3t^2-16t+25=0 are t= 8/3+i(sqrt(11))/3 ,t= 8/3-i(sqrt(11))/3

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3 t^{2} - 16 t + 25 = 0

t = \frac{8}{3} + i \frac{11}{3}, t = \frac{8}{3} - i \frac{11}{3}

Solution steps

3 t^{2} - 16 t + 25 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 3 \cdot 25}{2 \cdot 3}

Simplify

(- 16)^{2} - 4 \cdot 3 \cdot 25 : 211 i

t_{1, 2} = \frac{- ( - 16 ) \pm 2 11 i}{2 \cdot 3}

Separate the solutions

t_{1} = \frac{- ( - 16 ) + 2 11 i}{2 \cdot 3}, t_{2} = \frac{- ( - 16 ) - 2 11 i}{2 \cdot 3}

t = \frac{- ( - 16 ) + 2 11 i}{2 \cdot 3} : \frac{8}{3} + i \frac{11}{3}

t = \frac{- ( - 16 ) - 2 11 i}{2 \cdot 3} : \frac{8}{3} - i \frac{11}{3}

The solutions to the quadratic equation are:

t = \frac{8}{3} + i \frac{11}{3}, t = \frac{8}{3} - i \frac{11}{3}

\frac{( a x - 4 a - 2 b x + 8 b )}{( 2 a x + a y - 4 b x - 2 b y )}

\frac{x + 5}{1 - x} \leq \frac{3 - x}{x + 3}

(2 + 5 i) (5 - 3 i)

(\frac{2}{3})^{10}

x^{2} - 8 x - 33

What are the solutions to the equation 3t^2-16t+25=0 ?
The solutions to the equation 3t^2-16t+25=0 are t= 8/3+i(sqrt(11))/3 ,t= 8/3-i(sqrt(11))/3
Find the zeros of 3t^2-16t+25=0
The zeros of 3t^2-16t+25=0 are t= 8/3+i(sqrt(11))/3 ,t= 8/3-i(sqrt(11))/3