What are the solutions to the equation 6u^2+8u-1=0 ?

The solutions to the equation 6u^2+8u-1=0 are u=(-4+sqrt(22))/6 ,u=-(4+sqrt(22))/6

Find the zeros of 6u^2+8u-1=0

The zeros of 6u^2+8u-1=0 are u=(-4+sqrt(22))/6 ,u=-(4+sqrt(22))/6

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6 u^{2} + 8 u - 1 = 0

u = \frac{- 4 + 22}{6}, u = - \frac{4 + 22}{6}

Decimal Notation

u = 0.11506 \dots, u = - 1.44840 \dots

Solution steps

6 u^{2} + 8 u - 1 = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 6 ( - 1 )}{2 \cdot 6}

8^{2} - 4 \cdot 6 (- 1) = 222

u_{1, 2} = \frac{- 8 \pm 2 22}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- 8 + 2 22}{2 \cdot 6}, u_{2} = \frac{- 8 - 2 22}{2 \cdot 6}

u = \frac{- 8 + 2 22}{2 \cdot 6} : \frac{- 4 + 22}{6}

u = \frac{- 8 - 2 22}{2 \cdot 6} : - \frac{4 + 22}{6}

The solutions to the quadratic equation are:

u = \frac{- 4 + 22}{6}, u = - \frac{4 + 22}{6}

x^{2} + 8 x + 4 = 0

x (x + 3) - 2 x > 4 x + 4

\frac{11}{8} = \frac{k}{4}

(\frac{1}{3}) (\frac{2}{3})

x + 1 < 6

What are the solutions to the equation 6u^2+8u-1=0 ?
The solutions to the equation 6u^2+8u-1=0 are u=(-4+sqrt(22))/6 ,u=-(4+sqrt(22))/6
Find the zeros of 6u^2+8u-1=0
The zeros of 6u^2+8u-1=0 are u=(-4+sqrt(22))/6 ,u=-(4+sqrt(22))/6