What are the solutions to the equation 199-2x^2=x^2+33x+28 ?

The solutions to the equation 199-2x^2=x^2+33x+28 are x=(-11+sqrt(349))/2 ,x=-(11+sqrt(349))/2

Find the zeros of 199-2x^2=x^2+33x+28

The zeros of 199-2x^2=x^2+33x+28 are x=(-11+sqrt(349))/2 ,x=-(11+sqrt(349))/2

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199 - 2 x^{2} = x^{2} + 33 x + 28

x = \frac{- 11 + 349}{2}, x = - \frac{11 + 349}{2}

Solution steps

199 - 2 x^{2} = x^{2} + 33 x + 28

Switch sides

x^{2} + 33 x + 28 = 199 - 2 x^{2}

Move

2 x^{2}

to the left side

3 x^{2} + 33 x + 28 = 199

Move

199

to the left side

3 x^{2} + 33 x - 171 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 33 \pm 3 3 ^{2} - 4 \cdot 3 ( - 171 )}{2 \cdot 3}

3 3^{2} - 4 \cdot 3 (- 171) = 3349

x_{1, 2} = \frac{- 33 \pm 3 349}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 33 + 3 349}{2 \cdot 3}, x_{2} = \frac{- 33 - 3 349}{2 \cdot 3}

x = \frac{- 33 + 3 349}{2 \cdot 3} : \frac{- 11 + 349}{2}

x = \frac{- 33 - 3 349}{2 \cdot 3} : - \frac{11 + 349}{2}

The solutions to the quadratic equation are:

x = \frac{- 11 + 349}{2}, x = - \frac{11 + 349}{2}

6 m^{3} + 8 m^{2} + 9 m + 12

2 x^{2} - x - 3 < 0

x^{2} + (- \frac{2}{7})^{2} = 1

\frac{5 x + 5}{6} - 1 < \frac{19 x + 1}{3} - \frac{8 x + 1}{4}

- 2 x^{2} + 4 x

What are the solutions to the equation 199-2x^2=x^2+33x+28 ?
The solutions to the equation 199-2x^2=x^2+33x+28 are x=(-11+sqrt(349))/2 ,x=-(11+sqrt(349))/2
Find the zeros of 199-2x^2=x^2+33x+28
The zeros of 199-2x^2=x^2+33x+28 are x=(-11+sqrt(349))/2 ,x=-(11+sqrt(349))/2