What are the solutions to the equation x^2-4x+68=0 ?

The solutions to the equation x^2-4x+68=0 are x=2+8i,x=2-8i

Find the zeros of x^2-4x+68=0

The zeros of x^2-4x+68=0 are x=2+8i,x=2-8i

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x^{2} - 4 x + 68 = 0

x = 2 + 8 i, x = 2 - 8 i

Solution steps

x^{2} - 4 x + 68 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 68}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 68 : 16 i

x_{1, 2} = \frac{- ( - 4 ) \pm 16 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 16 i}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 16 i}{2 \cdot 1}

x = \frac{- ( - 4 ) + 16 i}{2 \cdot 1} : 2 + 8 i

x = \frac{- ( - 4 ) - 16 i}{2 \cdot 1} : 2 - 8 i

The solutions to the quadratic equation are:

x = 2 + 8 i, x = 2 - 8 i

3 (2 x - 1) = - 9 x

12 x^{2} - 75

1 < 2 - x < 2 x

x (x + 2) = 3

\frac{4000 π}{11}

What are the solutions to the equation x^2-4x+68=0 ?
The solutions to the equation x^2-4x+68=0 are x=2+8i,x=2-8i
Find the zeros of x^2-4x+68=0
The zeros of x^2-4x+68=0 are x=2+8i,x=2-8i