What are the solutions to the equation 18=-16x^2+16x+12 ?

The solutions to the equation 18=-16x^2+16x+12 are x= 1/2-i(sqrt(2))/4 ,x= 1/2+i(sqrt(2))/4

Find the zeros of 18=-16x^2+16x+12

The zeros of 18=-16x^2+16x+12 are x= 1/2-i(sqrt(2))/4 ,x= 1/2+i(sqrt(2))/4

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18 = - 16 x^{2} + 16 x + 12

x = \frac{1}{2} - i \frac{2}{4}, x = \frac{1}{2} + i \frac{2}{4}

Solution steps

18 = - 16 x^{2} + 16 x + 12

Switch sides

- 16 x^{2} + 16 x + 12 = 18

Move

18

to the left side

- 16 x^{2} + 16 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 16 ) ( - 6 )}{2 ( - 16 )}

Simplify

1 6^{2} - 4 (- 16) (- 6) : 82 i

x_{1, 2} = \frac{- 16 \pm 8 2 i}{2 ( - 16 )}

Separate the solutions

x_{1} = \frac{- 16 + 8 2 i}{2 ( - 16 )}, x_{2} = \frac{- 16 - 8 2 i}{2 ( - 16 )}

x = \frac{- 16 + 8 2 i}{2 ( - 16 )} : \frac{1}{2} - i \frac{2}{4}

x = \frac{- 16 - 8 2 i}{2 ( - 16 )} : \frac{1}{2} + i \frac{2}{4}

The solutions to the quadratic equation are:

x = \frac{1}{2} - i \frac{2}{4}, x = \frac{1}{2} + i \frac{2}{4}

7 (x^{2} - 9) = (x - 3)

4 x^{2} + 8 x - 320 = 0

(x - 2) (x + 4) - 2 (x - 4) = 0

3 y^{2} - 4 y + 2 = 2 (y - 3) 2

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

What are the solutions to the equation 18=-16x^2+16x+12 ?
The solutions to the equation 18=-16x^2+16x+12 are x= 1/2-i(sqrt(2))/4 ,x= 1/2+i(sqrt(2))/4
Find the zeros of 18=-16x^2+16x+12
The zeros of 18=-16x^2+16x+12 are x= 1/2-i(sqrt(2))/4 ,x= 1/2+i(sqrt(2))/4