What are the solutions to the equation 6+3x^2=-8x+2x^2-5 ?

The solutions to the equation 6+3x^2=-8x+2x^2-5 are x=-4-sqrt(5),x=-4+sqrt(5)

Find the zeros of 6+3x^2=-8x+2x^2-5

The zeros of 6+3x^2=-8x+2x^2-5 are x=-4-sqrt(5),x=-4+sqrt(5)

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6 + 3 x^{2} = - 8 x + 2 x^{2} - 5

x = - 4 - 5, x = - 4 + 5

Solution steps

6 + 3 x^{2} = - 8 x + 2 x^{2} - 5

Switch sides

- 8 x + 2 x^{2} - 5 = 6 + 3 x^{2}

Move

3 x^{2}

to the left side

- x^{2} - 8 x - 5 = 6

Move

6

to the left side

- x^{2} - 8 x - 11 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 1 ) ( - 11 )}{2 ( - 1 )}

(- 8)^{2} - 4 (- 1) (- 11) = 25

x_{1, 2} = \frac{- ( - 8 ) \pm 2 5}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 2 5}{2 ( - 1 )}, x_{2} = \frac{- ( - 8 ) - 2 5}{2 ( - 1 )}

x = \frac{- ( - 8 ) + 2 5}{2 ( - 1 )} : - 4 - 5

x = \frac{- ( - 8 ) - 2 5}{2 ( - 1 )} : - 4 + 5

The solutions to the quadratic equation are:

x = - 4 - 5, x = - 4 + 5

\frac{7}{a - 4} + \frac{2}{a}

5 (4 - 4 x)^{2} - 3 (4 - 4 x) (5 x - 1)

- 6 x + 8 = - 10 x - 20

- (- 3)^{3}

- 35 - 6 - 5

What are the solutions to the equation 6+3x^2=-8x+2x^2-5 ?
The solutions to the equation 6+3x^2=-8x+2x^2-5 are x=-4-sqrt(5),x=-4+sqrt(5)
Find the zeros of 6+3x^2=-8x+2x^2-5
The zeros of 6+3x^2=-8x+2x^2-5 are x=-4-sqrt(5),x=-4+sqrt(5)