What are the solutions to the equation t^2-4sqrt(3)t+8=0 ?

The solutions to the equation t^2-4sqrt(3)t+8=0 are t=2(1+sqrt(3)),t=2(sqrt(3)-1)

Find the zeros of t^2-4sqrt(3)t+8=0

The zeros of t^2-4sqrt(3)t+8=0 are t=2(1+sqrt(3)),t=2(sqrt(3)-1)

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t^{2} - 43 t + 8 = 0

t = 2 (1 + 3), t = 2 (3 - 1)

Decimal Notation

t = 5.46410 \dots, t = 1.46410 \dots

Solution steps

t^{2} - 43 t + 8 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 4 3 ) \pm ( - 4 3 ) ^{2} - 4 \cdot 1 \cdot 8}{2 \cdot 1}

(- 43)^{2} - 4 \cdot 1 \cdot 8 = 4

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

Separate the solutions

t_{1} = \frac{- ( - 4 3 ) + 4}{2 \cdot 1}, t_{2} = \frac{- ( - 4 3 ) - 4}{2 \cdot 1}

t = \frac{- ( - 4 3 ) + 4}{2 \cdot 1} : 2 (1 + 3)

t = \frac{- ( - 4 3 ) - 4}{2 \cdot 1} : 2 (3 - 1)

The solutions to the quadratic equation are:

t = 2 (1 + 3), t = 2 (3 - 1)

4 (a + 5) (a - 3) = (4 a + 20) (a - 3)

\frac{2 x ^{\frac{3}{2}}}{3}

(\frac{1}{3 a ^{3}})^{- 4}

- 6 a > 66

a^{2} - 2 a - 1 = 0

What are the solutions to the equation t^2-4sqrt(3)t+8=0 ?
The solutions to the equation t^2-4sqrt(3)t+8=0 are t=2(1+sqrt(3)),t=2(sqrt(3)-1)
Find the zeros of t^2-4sqrt(3)t+8=0
The zeros of t^2-4sqrt(3)t+8=0 are t=2(1+sqrt(3)),t=2(sqrt(3)-1)