What are the solutions to the equation 2x^2+12x+8=8x+4 ?

The solutions to the equation 2x^2+12x+8=8x+4 are x=-1+i,x=-1-i

Find the zeros of 2x^2+12x+8=8x+4

The zeros of 2x^2+12x+8=8x+4 are x=-1+i,x=-1-i

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2 x^{2} + 12 x + 8 = 8 x + 4

x = - 1 + i, x = - 1 - i

Solution steps

2 x^{2} + 12 x + 8 = 8 x + 4

Move

4

to the left side

2 x^{2} + 12 x + 4 = 8 x

Move

8 x

to the left side

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

Solve with the quadratic formula

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

Simplify

4^{2} - 4 \cdot 2 \cdot 4 : 4 i

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

Separate the solutions

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

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

x = \frac{- 4 - 4 i}{2 \cdot 2} : - 1 - i

The solutions to the quadratic equation are:

x = - 1 + i, x = - 1 - i

\frac{- 5 + ( 5 ) ^{2} - 4 ( 3 ) ( - 2 )}{2 ( 3 )} = x

3 x + 2 < 2 x - 4

2 r^{2} + 5 r - 3 = 0

(7 x - 3 y) (2 x - 3 y) - (3 y - 2 x)^{2}

What are the solutions to the equation 2x^2+12x+8=8x+4 ?
The solutions to the equation 2x^2+12x+8=8x+4 are x=-1+i,x=-1-i
Find the zeros of 2x^2+12x+8=8x+4
The zeros of 2x^2+12x+8=8x+4 are x=-1+i,x=-1-i