What are the solutions to the equation 6x^2+7x=11 ?

The solutions to the equation 6x^2+7x=11 are x=(-7+sqrt(313))/(12),x=(-7-sqrt(313))/(12)

Find the zeros of 6x^2+7x=11

The zeros of 6x^2+7x=11 are x=(-7+sqrt(313))/(12),x=(-7-sqrt(313))/(12)

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6 x^{2} + 7 x = 11

x = \frac{- 7 + 313}{12}, x = \frac{- 7 - 313}{12}

Decimal Notation

x = 0.89098 \dots, x = - 2.05765 \dots

Solution steps

6 x^{2} + 7 x = 11

Move

11

to the left side

6 x^{2} + 7 x - 11 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 6 ( - 11 )}{2 \cdot 6}

7^{2} - 4 \cdot 6 (- 11) = 313

x_{1, 2} = \frac{- 7 \pm 313}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 7 + 313}{2 \cdot 6}, x_{2} = \frac{- 7 - 313}{2 \cdot 6}

x = \frac{- 7 + 313}{2 \cdot 6} : \frac{- 7 + 313}{12}

x = \frac{- 7 - 313}{2 \cdot 6} : \frac{- 7 - 313}{12}

The solutions to the quadratic equation are:

x = \frac{- 7 + 313}{12}, x = \frac{- 7 - 313}{12}

(x - 4)^{2} - 10 (x - 4) + 24 = 0

(13 m^{4} + 2) + (m^{4} n^{2} + 2 - 2 m^{4}) - (- 13 m^{2} n^{3} + 5 m^{4})

22500 - 740 x = 15200 + 1500 x

(x + 8)^{2} - 4

\frac{x}{4} = 12

What are the solutions to the equation 6x^2+7x=11 ?
The solutions to the equation 6x^2+7x=11 are x=(-7+sqrt(313))/(12),x=(-7-sqrt(313))/(12)
Find the zeros of 6x^2+7x=11
The zeros of 6x^2+7x=11 are x=(-7+sqrt(313))/(12),x=(-7-sqrt(313))/(12)