What are the solutions to the equation 3t^2-12t+11=0 ?

The solutions to the equation 3t^2-12t+11=0 are t=(6+sqrt(3))/3 ,t=(6-sqrt(3))/3

Find the zeros of 3t^2-12t+11=0

The zeros of 3t^2-12t+11=0 are t=(6+sqrt(3))/3 ,t=(6-sqrt(3))/3

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3 t^{2} - 12 t + 11 = 0

t = \frac{6 + 3}{3}, t = \frac{6 - 3}{3}

Decimal Notation

t = 2.57735 \dots, t = 1.42264 \dots

Solution steps

3 t^{2} - 12 t + 11 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 3 \cdot 11}{2 \cdot 3}

(- 12)^{2} - 4 \cdot 3 \cdot 11 = 23

t_{1, 2} = \frac{- ( - 12 ) \pm 2 3}{2 \cdot 3}

Separate the solutions

t_{1} = \frac{- ( - 12 ) + 2 3}{2 \cdot 3}, t_{2} = \frac{- ( - 12 ) - 2 3}{2 \cdot 3}

t = \frac{- ( - 12 ) + 2 3}{2 \cdot 3} : \frac{6 + 3}{3}

t = \frac{- ( - 12 ) - 2 3}{2 \cdot 3} : \frac{6 - 3}{3}

The solutions to the quadratic equation are:

t = \frac{6 + 3}{3}, t = \frac{6 - 3}{3}

x^{2} - 3 x + x y - 3 y

(9 - 8 i) - (- 6 - i)

2 x^{2} - 3 x y

x^{5} - 4 x^{4} - x^{3} + 10 x^{2} + 2 x - 4

4 ∣ t + 1 ∣ \geq 16

What are the solutions to the equation 3t^2-12t+11=0 ?
The solutions to the equation 3t^2-12t+11=0 are t=(6+sqrt(3))/3 ,t=(6-sqrt(3))/3
Find the zeros of 3t^2-12t+11=0
The zeros of 3t^2-12t+11=0 are t=(6+sqrt(3))/3 ,t=(6-sqrt(3))/3