What are the solutions to the equation x^2-8x=4x-38 ?

The solutions to the equation x^2-8x=4x-38 are x=6+sqrt(2)i,x=6-sqrt(2)i

Find the zeros of x^2-8x=4x-38

The zeros of x^2-8x=4x-38 are x=6+sqrt(2)i,x=6-sqrt(2)i

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x^{2} - 8 x = 4 x - 38

x = 6 + 2 i, x = 6 - 2 i

Solution steps

x^{2} - 8 x = 4 x - 38

Move

38

to the left side

x^{2} - 8 x + 38 = 4 x

Move

4 x

to the left side

x^{2} - 12 x + 38 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 38}{2 \cdot 1}

Simplify

(- 12)^{2} - 4 \cdot 1 \cdot 38 : 22 i

x_{1, 2} = \frac{- ( - 12 ) \pm 2 2 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 2 2 i}{2 \cdot 1}, x_{2} = \frac{- ( - 12 ) - 2 2 i}{2 \cdot 1}

x = \frac{- ( - 12 ) + 2 2 i}{2 \cdot 1} : 6 + 2 i

x = \frac{- ( - 12 ) - 2 2 i}{2 \cdot 1} : 6 - 2 i

The solutions to the quadratic equation are:

x = 6 + 2 i, x = 6 - 2 i

y^{2} - 16 y + 64 = 0

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

x^{2} + 2 x + 1 = 0

4 (6 - 3 x)^{2} - 38 = - 37

m^{2} + 10 m + 25 = 0

What are the solutions to the equation x^2-8x=4x-38 ?
The solutions to the equation x^2-8x=4x-38 are x=6+sqrt(2)i,x=6-sqrt(2)i
Find the zeros of x^2-8x=4x-38
The zeros of x^2-8x=4x-38 are x=6+sqrt(2)i,x=6-sqrt(2)i