What are the solutions to the equation (x^2-x+5)/4 =0 ?

The solutions to the equation (x^2-x+5)/4 =0 are x= 1/2+i(sqrt(19))/2 ,x= 1/2-i(sqrt(19))/2

Find the zeros of (x^2-x+5)/4 =0

The zeros of (x^2-x+5)/4 =0 are x= 1/2+i(sqrt(19))/2 ,x= 1/2-i(sqrt(19))/2

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\frac{x ^{2} - x + 5}{4} = 0

x = \frac{1}{2} + i \frac{19}{2}, x = \frac{1}{2} - i \frac{19}{2}

Solution steps

\frac{x ^{2} - x + 5}{4} = 0

Multiply both sides by

4

x^{2} - x + 5 = 0

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 1 \cdot 5 : 19 i

x_{1, 2} = \frac{- ( - 1 ) \pm 19 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 19 i}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 19 i}{2 \cdot 1}

x = \frac{- ( - 1 ) + 19 i}{2 \cdot 1} : \frac{1}{2} + i \frac{19}{2}

x = \frac{- ( - 1 ) - 19 i}{2 \cdot 1} : \frac{1}{2} - i \frac{19}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{19}{2}, x = \frac{1}{2} - i \frac{19}{2}

- 63 = - 2 y^{2} + 9

x, 16 x^{2} - 24 x y + 9 y^{2} - 60 x - 80 y + 100 = 0

(x + 1) (x + 2) - (x - 3) (+ 4) = 6

x^{2} + x + 1 = 7

n (25 - x^{2}) = 84

What are the solutions to the equation (x^2-x+5)/4 =0 ?
The solutions to the equation (x^2-x+5)/4 =0 are x= 1/2+i(sqrt(19))/2 ,x= 1/2-i(sqrt(19))/2
Find the zeros of (x^2-x+5)/4 =0
The zeros of (x^2-x+5)/4 =0 are x= 1/2+i(sqrt(19))/2 ,x= 1/2-i(sqrt(19))/2