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

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

Find the zeros of (x)=2x^2+3x+1

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

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(x) = 2 x^{2} + 3 x + 1

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

Solution steps

(x) = 2 x^{2} + 3 x + 1

Refine

x = 2 x^{2} + 3 x + 1

Switch sides

2 x^{2} + 3 x + 1 = x

Move

x

to the left side

2 x^{2} + 2 x + 1 = 0

Solve with the quadratic formula

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

Simplify

2^{2} - 4 \cdot 2 \cdot 1 : 2 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

2 x^{2} + 4 = 9 x

x^{2} + x - 2 - 108 = 0

x (x - 1) + 2 (x - 2) = - 2

12 a^{2} + 36 a = 0

v^{2} - 81 = 0

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