What are the solutions to the equation 4x^2+5x+2=2x^2+7x-1 ?

The solutions to the equation 4x^2+5x+2=2x^2+7x-1 are x= 1/2+i(sqrt(5))/2 ,x= 1/2-i(sqrt(5))/2

Find the zeros of 4x^2+5x+2=2x^2+7x-1

The zeros of 4x^2+5x+2=2x^2+7x-1 are x= 1/2+i(sqrt(5))/2 ,x= 1/2-i(sqrt(5))/2

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4 x^{2} + 5 x + 2 = 2 x^{2} + 7 x - 1

x = \frac{1}{2} + i \frac{5}{2}, x = \frac{1}{2} - i \frac{5}{2}

Solution steps

4 x^{2} + 5 x + 2 = 2 x^{2} + 7 x - 1

Move

1

to the left side

4 x^{2} + 5 x + 3 = 2 x^{2} + 7 x

Move

7 x

to the left side

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

Move

2 x^{2}

to the left side

2 x^{2} - 2 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

Simplify

(- 2)^{2} - 4 \cdot 2 \cdot 3 : 25 i

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

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 5 i}{2 \cdot 2}, x_{2} = \frac{- ( - 2 ) - 2 5 i}{2 \cdot 2}

x = \frac{- ( - 2 ) + 2 5 i}{2 \cdot 2} : \frac{1}{2} + i \frac{5}{2}

x = \frac{- ( - 2 ) - 2 5 i}{2 \cdot 2} : \frac{1}{2} - i \frac{5}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{5}{2}, x = \frac{1}{2} - i \frac{5}{2}

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

w^{2} + 3 w - 4 = 0

r^{2} - 7 r = 0

9000 = 4000 (1 + r)^{2}

5 - 5

What are the solutions to the equation 4x^2+5x+2=2x^2+7x-1 ?
The solutions to the equation 4x^2+5x+2=2x^2+7x-1 are x= 1/2+i(sqrt(5))/2 ,x= 1/2-i(sqrt(5))/2
Find the zeros of 4x^2+5x+2=2x^2+7x-1
The zeros of 4x^2+5x+2=2x^2+7x-1 are x= 1/2+i(sqrt(5))/2 ,x= 1/2-i(sqrt(5))/2