What are the solutions to the equation 3x^2-68=2x^2+6x+11 ?

The solutions to the equation 3x^2-68=2x^2+6x+11 are x=3-2sqrt(22),x=3+2sqrt(22)

Find the zeros of 3x^2-68=2x^2+6x+11

The zeros of 3x^2-68=2x^2+6x+11 are x=3-2sqrt(22),x=3+2sqrt(22)

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3 x^{2} - 68 = 2 x^{2} + 6 x + 11

x = 3 - 222, x = 3 + 222

Solution steps

3 x^{2} - 68 = 2 x^{2} + 6 x + 11

Switch sides

2 x^{2} + 6 x + 11 = 3 x^{2} - 68

Move

68

to the left side

2 x^{2} + 6 x + 79 = 3 x^{2}

Move

3 x^{2}

to the left side

- x^{2} + 6 x + 79 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 1 ) \cdot 79}{2 ( - 1 )}

6^{2} - 4 (- 1) \cdot 79 = 422

x_{1, 2} = \frac{- 6 \pm 4 22}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 6 + 4 22}{2 ( - 1 )}, x_{2} = \frac{- 6 - 4 22}{2 ( - 1 )}

x = \frac{- 6 + 4 22}{2 ( - 1 )} : 3 - 222

x = \frac{- 6 - 4 22}{2 ( - 1 )} : 3 + 222

The solutions to the quadratic equation are:

x = 3 - 222, x = 3 + 222

z^{2} - 14 z + 49 = - 121

3 n^{2} + 5 = - 46

x^{2} + 15 x - 23 = 0

5 t^{2} + 60 = 0

x^{2} + x + 2.5 = 0

What are the solutions to the equation 3x^2-68=2x^2+6x+11 ?
The solutions to the equation 3x^2-68=2x^2+6x+11 are x=3-2sqrt(22),x=3+2sqrt(22)
Find the zeros of 3x^2-68=2x^2+6x+11
The zeros of 3x^2-68=2x^2+6x+11 are x=3-2sqrt(22),x=3+2sqrt(22)