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

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

Find the zeros of (x)=2x^2+3x+4

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

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(x) = 2 x^{2} + 3 x + 4

x = - \frac{1}{2} + i \frac{7}{2}, x = - \frac{1}{2} - i \frac{7}{2}

Solution steps

(x) = 2 x^{2} + 3 x + 4

Refine

x = 2 x^{2} + 3 x + 4

Switch sides

2 x^{2} + 3 x + 4 = x

Move

x

to the left side

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

Solve with the quadratic formula

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

Simplify

2^{2} - 4 \cdot 2 \cdot 4 : 27 i

x_{1, 2} = \frac{- 2 \pm 2 7 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 2 + 2 7 i}{2 \cdot 2}, x_{2} = \frac{- 2 - 2 7 i}{2 \cdot 2}

x = \frac{- 2 + 2 7 i}{2 \cdot 2} : - \frac{1}{2} + i \frac{7}{2}

x = \frac{- 2 - 2 7 i}{2 \cdot 2} : - \frac{1}{2} - i \frac{7}{2}

The solutions to the quadratic equation are:

x = - \frac{1}{2} + i \frac{7}{2}, x = - \frac{1}{2} - i \frac{7}{2}

15 - 2 x^{2} = 3 x

4 x^{2} - 16 x + 144 = 0

x^{2} + 8 x + 19

(y + 2) (y - 3) = - 6

3 x - 4 = 5 x^{2}

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