What are the solutions to the equation x^2-3x=-2x-7 ?

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

Find the zeros of x^2-3x=-2x-7

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

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x^{2} - 3 x = - 2 x - 7

x = \frac{1}{2} + i \frac{3 3}{2}, x = \frac{1}{2} - i \frac{3 3}{2}

Solution steps

x^{2} - 3 x = - 2 x - 7

Move

7

to the left side

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

Move

2 x

to the left side

x^{2} - x + 7 = 0

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 1 \cdot 7 : 33 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{3 3}{2}, x = \frac{1}{2} - i \frac{3 3}{2}

2 x^{2} - 6 x + n - 1 = 0

15 r^{2} - 16 r - 20 = - 5

x^{2} + 4 x - 14 = 4 x - 5

x, f = 4 x^{2} y^{2} - 5 x y^{3} + 12

5 - 2 (3 x^{2} - 2 x) = x^{2} + 7 - 2 x^{2}

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