What are the solutions to the equation (x^2+4)-(x-2)(3-2x)=0 ?

The solutions to the equation (x^2+4)-(x-2)(3-2x)=0 are x= 7/6+i(sqrt(71))/6 ,x= 7/6-i(sqrt(71))/6

Find the zeros of (x^2+4)-(x-2)(3-2x)=0

The zeros of (x^2+4)-(x-2)(3-2x)=0 are x= 7/6+i(sqrt(71))/6 ,x= 7/6-i(sqrt(71))/6

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(x^{2} + 4) - (x - 2) (3 - 2 x) = 0

x = \frac{7}{6} + i \frac{71}{6}, x = \frac{7}{6} - i \frac{71}{6}

Solution steps

(x^{2} + 4) - (x - 2) (3 - 2 x) = 0

Expand

(x^{2} + 4) - (x - 2) (3 - 2 x) : 3 x^{2} - 7 x + 10

3 x^{2} - 7 x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 3 \cdot 10}{2 \cdot 3}

Simplify

(- 7)^{2} - 4 \cdot 3 \cdot 10 : 71 i

x_{1, 2} = \frac{- ( - 7 ) \pm 71 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 71 i}{2 \cdot 3}, x_{2} = \frac{- ( - 7 ) - 71 i}{2 \cdot 3}

x = \frac{- ( - 7 ) + 71 i}{2 \cdot 3} : \frac{7}{6} + i \frac{71}{6}

x = \frac{- ( - 7 ) - 71 i}{2 \cdot 3} : \frac{7}{6} - i \frac{71}{6}

The solutions to the quadratic equation are:

x = \frac{7}{6} + i \frac{71}{6}, x = \frac{7}{6} - i \frac{71}{6}

x^{2} - 8 x + 11 = 1

x^{2} - 6 x + 9 = 3

(2 x - 1) (3 x + 7) = 0

4 x^{2} - 9 x - 36 = 0

x^{2} + (- 2) x - 8 = 0

What are the solutions to the equation (x^2+4)-(x-2)(3-2x)=0 ?
The solutions to the equation (x^2+4)-(x-2)(3-2x)=0 are x= 7/6+i(sqrt(71))/6 ,x= 7/6-i(sqrt(71))/6
Find the zeros of (x^2+4)-(x-2)(3-2x)=0
The zeros of (x^2+4)-(x-2)(3-2x)=0 are x= 7/6+i(sqrt(71))/6 ,x= 7/6-i(sqrt(71))/6