What are the solutions to the equation 11x^2+7x=3 ?

The solutions to the equation 11x^2+7x=3 are x=(-7+sqrt(181))/(22),x=(-7-sqrt(181))/(22)

Find the zeros of 11x^2+7x=3

The zeros of 11x^2+7x=3 are x=(-7+sqrt(181))/(22),x=(-7-sqrt(181))/(22)

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11 x^{2} + 7 x = 3

x = \frac{- 7 + 181}{22}, x = \frac{- 7 - 181}{22}

Decimal Notation

x = 0.29334 \dots, x = - 0.92971 \dots

Solution steps

11 x^{2} + 7 x = 3

Move

3

to the left side

11 x^{2} + 7 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 11 ( - 3 )}{2 \cdot 11}

7^{2} - 4 \cdot 11 (- 3) = 181

x_{1, 2} = \frac{- 7 \pm 181}{2 \cdot 11}

Separate the solutions

x_{1} = \frac{- 7 + 181}{2 \cdot 11}, x_{2} = \frac{- 7 - 181}{2 \cdot 11}

x = \frac{- 7 + 181}{2 \cdot 11} : \frac{- 7 + 181}{22}

x = \frac{- 7 - 181}{2 \cdot 11} : \frac{- 7 - 181}{22}

The solutions to the quadratic equation are:

x = \frac{- 7 + 181}{22}, x = \frac{- 7 - 181}{22}

0.69 x^{2} - 34 x + 435 = 600

3 x^{2} = - 4 + 8 x

121 x^{2} - 64 = 0

r^{2} + 12 r + 17 = 0

x^{2} - x + 4

What are the solutions to the equation 11x^2+7x=3 ?
The solutions to the equation 11x^2+7x=3 are x=(-7+sqrt(181))/(22),x=(-7-sqrt(181))/(22)
Find the zeros of 11x^2+7x=3
The zeros of 11x^2+7x=3 are x=(-7+sqrt(181))/(22),x=(-7-sqrt(181))/(22)