What are the solutions to the equation 12-4x+2x^2=7 ?

The solutions to the equation 12-4x+2x^2=7 are x=1+isqrt(3/2),x=1-isqrt(3/2)

Find the zeros of 12-4x+2x^2=7

The zeros of 12-4x+2x^2=7 are x=1+isqrt(3/2),x=1-isqrt(3/2)

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12 - 4 x + 2 x^{2} = 7

x = 1 + i \frac{3}{2}, x = 1 - i \frac{3}{2}

Solution steps

12 - 4 x + 2 x^{2} = 7

Move

7

to the left side

2 x^{2} - 4 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

Simplify

(- 4)^{2} - 4 \cdot 2 \cdot 5 : 26 i

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

Separate the solutions

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

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

x = \frac{- ( - 4 ) - 2 6 i}{2 \cdot 2} : 1 - i \frac{3}{2}

The solutions to the quadratic equation are:

x = 1 + i \frac{3}{2}, x = 1 - i \frac{3}{2}

3 r^{2} = 6 r + 5

(x - 3) (x - 1) = 8

5 x^{2} = 17

- 4 x^{2} + x - 25 = - 3 x^{2} + 17 x + 38

(3 y - 1) (y + 2) = 0

What are the solutions to the equation 12-4x+2x^2=7 ?
The solutions to the equation 12-4x+2x^2=7 are x=1+isqrt(3/2),x=1-isqrt(3/2)
Find the zeros of 12-4x+2x^2=7
The zeros of 12-4x+2x^2=7 are x=1+isqrt(3/2),x=1-isqrt(3/2)