What are the solutions to the equation 3(4x^2-1)+6(-2x+3)=-7 ?

The solutions to the equation 3(4x^2-1)+6(-2x+3)=-7 are x= 1/2+i(sqrt(57))/6 ,x= 1/2-i(sqrt(57))/6

Find the zeros of 3(4x^2-1)+6(-2x+3)=-7

The zeros of 3(4x^2-1)+6(-2x+3)=-7 are x= 1/2+i(sqrt(57))/6 ,x= 1/2-i(sqrt(57))/6

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3 (4 x^{2} - 1) + 6 (- 2 x + 3) = - 7

x = \frac{1}{2} + i \frac{57}{6}, x = \frac{1}{2} - i \frac{57}{6}

Solution steps

3 (4 x^{2} - 1) + 6 (- 2 x + 3) = - 7

Expand

3 (4 x^{2} - 1) + 6 (- 2 x + 3) : 12 x^{2} - 12 x + 15

12 x^{2} - 12 x + 15 = - 7

Move

7

to the left side

12 x^{2} - 12 x + 22 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 12 \cdot 22}{2 \cdot 12}

Simplify

(- 12)^{2} - 4 \cdot 12 \cdot 22 : 457 i

x_{1, 2} = \frac{- ( - 12 ) \pm 4 57 i}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 4 57 i}{2 \cdot 12}, x_{2} = \frac{- ( - 12 ) - 4 57 i}{2 \cdot 12}

x = \frac{- ( - 12 ) + 4 57 i}{2 \cdot 12} : \frac{1}{2} + i \frac{57}{6}

x = \frac{- ( - 12 ) - 4 57 i}{2 \cdot 12} : \frac{1}{2} - i \frac{57}{6}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{57}{6}, x = \frac{1}{2} - i \frac{57}{6}

5 x^{2} - 2 x - 35 = 0

4 n^{2} - 20 n + 16 = 0

220 - (16.5 - 2 x)^{2} = 99.9

5 x^{2} + 20 x

8 z^{2} - 3 z - 2 = 0

What are the solutions to the equation 3(4x^2-1)+6(-2x+3)=-7 ?
The solutions to the equation 3(4x^2-1)+6(-2x+3)=-7 are x= 1/2+i(sqrt(57))/6 ,x= 1/2-i(sqrt(57))/6
Find the zeros of 3(4x^2-1)+6(-2x+3)=-7
The zeros of 3(4x^2-1)+6(-2x+3)=-7 are x= 1/2+i(sqrt(57))/6 ,x= 1/2-i(sqrt(57))/6