What are the solutions to the equation x^2-3x+5=2x-9 ?

The solutions to the equation x^2-3x+5=2x-9 are x= 5/2+i(sqrt(31))/2 ,x= 5/2-i(sqrt(31))/2

Find the zeros of x^2-3x+5=2x-9

The zeros of x^2-3x+5=2x-9 are x= 5/2+i(sqrt(31))/2 ,x= 5/2-i(sqrt(31))/2

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x^{2} - 3 x + 5 = 2 x - 9

x = \frac{5}{2} + i \frac{31}{2}, x = \frac{5}{2} - i \frac{31}{2}

Solution steps

x^{2} - 3 x + 5 = 2 x - 9

Move

9

to the left side

x^{2} - 3 x + 14 = 2 x

Move

2 x

to the left side

x^{2} - 5 x + 14 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 14}{2 \cdot 1}

Simplify

(- 5)^{2} - 4 \cdot 1 \cdot 14 : 31 i

x_{1, 2} = \frac{- ( - 5 ) \pm 31 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 31 i}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 31 i}{2 \cdot 1}

x = \frac{- ( - 5 ) + 31 i}{2 \cdot 1} : \frac{5}{2} + i \frac{31}{2}

x = \frac{- ( - 5 ) - 31 i}{2 \cdot 1} : \frac{5}{2} - i \frac{31}{2}

The solutions to the quadratic equation are:

x = \frac{5}{2} + i \frac{31}{2}, x = \frac{5}{2} - i \frac{31}{2}

2 (5 x + 9)^{2} + 10 = 28

2 x^{2} + 5 = 5 x^{2} - 37

(x + 3)^{2} - 3 = 49

4 (x^{2} + 6) + 35 x = 0

(x - 7)^{2} + (x + 1)^{2} = 7 (x - 1)^{2} + 43

What are the solutions to the equation x^2-3x+5=2x-9 ?
The solutions to the equation x^2-3x+5=2x-9 are x= 5/2+i(sqrt(31))/2 ,x= 5/2-i(sqrt(31))/2
Find the zeros of x^2-3x+5=2x-9
The zeros of x^2-3x+5=2x-9 are x= 5/2+i(sqrt(31))/2 ,x= 5/2-i(sqrt(31))/2