What are the solutions to the equation -3x+19=-x^2+1 ?

The solutions to the equation -3x+19=-x^2+1 are x= 3/2-i(3sqrt(7))/2 ,x= 3/2+i(3sqrt(7))/2

Find the zeros of -3x+19=-x^2+1

The zeros of -3x+19=-x^2+1 are x= 3/2-i(3sqrt(7))/2 ,x= 3/2+i(3sqrt(7))/2

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- 3 x + 19 = - x^{2} + 1

x = \frac{3}{2} - i \frac{3 7}{2}, x = \frac{3}{2} + i \frac{3 7}{2}

Solution steps

- 3 x + 19 = - x^{2} + 1

Switch sides

- x^{2} + 1 = - 3 x + 19

Move

19

to the left side

- x^{2} - 18 = - 3 x

Move

3 x

to the left side

- x^{2} - 18 + 3 x = 0

Write in the standard form

a x^{2} + b x + c = 0

- x^{2} + 3 x - 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 1 ) ( - 18 )}{2 ( - 1 )}

Simplify

3^{2} - 4 (- 1) (- 18) : 37 i

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

Separate the solutions

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

x = \frac{- 3 + 3 7 i}{2 ( - 1 )} : \frac{3}{2} - i \frac{3 7}{2}

x = \frac{- 3 - 3 7 i}{2 ( - 1 )} : \frac{3}{2} + i \frac{3 7}{2}

The solutions to the quadratic equation are:

x = \frac{3}{2} - i \frac{3 7}{2}, x = \frac{3}{2} + i \frac{3 7}{2}

50 + 18

x^{3} - x^{2} - 30 x

\frac{9 + 4 i}{6 i}

\frac{1}{4} \cdot 3

What are the solutions to the equation -3x+19=-x^2+1 ?
The solutions to the equation -3x+19=-x^2+1 are x= 3/2-i(3sqrt(7))/2 ,x= 3/2+i(3sqrt(7))/2
Find the zeros of -3x+19=-x^2+1
The zeros of -3x+19=-x^2+1 are x= 3/2-i(3sqrt(7))/2 ,x= 3/2+i(3sqrt(7))/2