What are the solutions to the equation (2x-7)(x+3)+30=x(1-x) ?

The solutions to the equation (2x-7)(x+3)+30=x(1-x) are x= 1/3+i(sqrt(26))/3 ,x= 1/3-i(sqrt(26))/3

Find the zeros of (2x-7)(x+3)+30=x(1-x)

The zeros of (2x-7)(x+3)+30=x(1-x) are x= 1/3+i(sqrt(26))/3 ,x= 1/3-i(sqrt(26))/3

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(2 x - 7) (x + 3) + 30 = x (1 - x)

x = \frac{1}{3} + i \frac{26}{3}, x = \frac{1}{3} - i \frac{26}{3}

Solution steps

(2 x - 7) (x + 3) + 30 = x (1 - x)

Expand

(2 x - 7) (x + 3) + 30 : 2 x^{2} - x + 9

Expand

x (1 - x) : x - x^{2}

2 x^{2} - x + 9 = x - x^{2}

Move

x^{2}

to the left side

3 x^{2} - x + 9 = x

Move

x

to the left side

3 x^{2} - 2 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 3 \cdot 9}{2 \cdot 3}

Simplify

(- 2)^{2} - 4 \cdot 3 \cdot 9 : 226 i

x_{1, 2} = \frac{- ( - 2 ) \pm 2 26 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 26 i}{2 \cdot 3}, x_{2} = \frac{- ( - 2 ) - 2 26 i}{2 \cdot 3}

x = \frac{- ( - 2 ) + 2 26 i}{2 \cdot 3} : \frac{1}{3} + i \frac{26}{3}

x = \frac{- ( - 2 ) - 2 26 i}{2 \cdot 3} : \frac{1}{3} - i \frac{26}{3}

The solutions to the quadratic equation are:

x = \frac{1}{3} + i \frac{26}{3}, x = \frac{1}{3} - i \frac{26}{3}

64 = 2 x^{2}

4 x^{2} + 28 x + 49 = 0

x^{2} + 100 x + 10000 = 0

x, u = x^{2} y

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

What are the solutions to the equation (2x-7)(x+3)+30=x(1-x) ?
The solutions to the equation (2x-7)(x+3)+30=x(1-x) are x= 1/3+i(sqrt(26))/3 ,x= 1/3-i(sqrt(26))/3
Find the zeros of (2x-7)(x+3)+30=x(1-x)
The zeros of (2x-7)(x+3)+30=x(1-x) are x= 1/3+i(sqrt(26))/3 ,x= 1/3-i(sqrt(26))/3