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

The solutions to the equation 4x^2-x+5=0 are x= 1/8+i(sqrt(79))/8 ,x= 1/8-i(sqrt(79))/8

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

The zeros of 4x^2-x+5=0 are x= 1/8+i(sqrt(79))/8 ,x= 1/8-i(sqrt(79))/8

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

x = \frac{1}{8} + i \frac{79}{8}, x = \frac{1}{8} - i \frac{79}{8}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 4 \cdot 5 : 79 i

x_{1, 2} = \frac{- ( - 1 ) \pm 79 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 79 i}{2 \cdot 4}, x_{2} = \frac{- ( - 1 ) - 79 i}{2 \cdot 4}

x = \frac{- ( - 1 ) + 79 i}{2 \cdot 4} : \frac{1}{8} + i \frac{79}{8}

x = \frac{- ( - 1 ) - 79 i}{2 \cdot 4} : \frac{1}{8} - i \frac{79}{8}

The solutions to the quadratic equation are:

x = \frac{1}{8} + i \frac{79}{8}, x = \frac{1}{8} - i \frac{79}{8}

\frac{2 x ^{2} - 6}{2 ( x + 1 )} - \frac{4 x}{2 x + 2} < 4 x + 1

x (x + 4) = 45

(x^{3})^{- 3} (- 2 y^{5})^{4}

t^{2} + 9 t - 4 = - 5

4 x^{2} - 23 x + 15

What are the solutions to the equation 4x^2-x+5=0 ?
The solutions to the equation 4x^2-x+5=0 are x= 1/8+i(sqrt(79))/8 ,x= 1/8-i(sqrt(79))/8
Find the zeros of 4x^2-x+5=0
The zeros of 4x^2-x+5=0 are x= 1/8+i(sqrt(79))/8 ,x= 1/8-i(sqrt(79))/8