What are the solutions to the equation x^2+2x-4=3x-8 ?

The solutions to the equation x^2+2x-4=3x-8 are x= 1/2+i(sqrt(15))/2 ,x= 1/2-i(sqrt(15))/2

Find the zeros of x^2+2x-4=3x-8

The zeros of x^2+2x-4=3x-8 are x= 1/2+i(sqrt(15))/2 ,x= 1/2-i(sqrt(15))/2

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x^{2} + 2 x - 4 = 3 x - 8

x = \frac{1}{2} + i \frac{15}{2}, x = \frac{1}{2} - i \frac{15}{2}

Solution steps

x^{2} + 2 x - 4 = 3 x - 8

Move

8

to the left side

x^{2} + 2 x + 4 = 3 x

Move

3 x

to the left side

x^{2} - x + 4 = 0

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 1 \cdot 4 : 15 i

x_{1, 2} = \frac{- ( - 1 ) \pm 15 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 15 i}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 15 i}{2 \cdot 1}

x = \frac{- ( - 1 ) + 15 i}{2 \cdot 1} : \frac{1}{2} + i \frac{15}{2}

x = \frac{- ( - 1 ) - 15 i}{2 \cdot 1} : \frac{1}{2} - i \frac{15}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{15}{2}, x = \frac{1}{2} - i \frac{15}{2}

9^{2} + 4 0^{2} = c^{2}

4 x^{2} - 9 = 7

- 4 x^{2} + 64 x + 256 = 0

x^{2} - 17 x + 52 = 0

- \frac{1}{2} x^{2} - x + \frac{3}{2} = 0

What are the solutions to the equation x^2+2x-4=3x-8 ?
The solutions to the equation x^2+2x-4=3x-8 are x= 1/2+i(sqrt(15))/2 ,x= 1/2-i(sqrt(15))/2
Find the zeros of x^2+2x-4=3x-8
The zeros of x^2+2x-4=3x-8 are x= 1/2+i(sqrt(15))/2 ,x= 1/2-i(sqrt(15))/2