What are the solutions to the equation 8x^2-4x+21=0 ?

The solutions to the equation 8x^2-4x+21=0 are x= 1/4+i(sqrt(41))/4 ,x= 1/4-i(sqrt(41))/4

Find the zeros of 8x^2-4x+21=0

The zeros of 8x^2-4x+21=0 are x= 1/4+i(sqrt(41))/4 ,x= 1/4-i(sqrt(41))/4

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8 x^{2} - 4 x + 21 = 0

x = \frac{1}{4} + i \frac{41}{4}, x = \frac{1}{4} - i \frac{41}{4}

Solution steps

8 x^{2} - 4 x + 21 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 8 \cdot 21}{2 \cdot 8}

Simplify

(- 4)^{2} - 4 \cdot 8 \cdot 21 : 441 i

x_{1, 2} = \frac{- ( - 4 ) \pm 4 41 i}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 41 i}{2 \cdot 8}, x_{2} = \frac{- ( - 4 ) - 4 41 i}{2 \cdot 8}

x = \frac{- ( - 4 ) + 4 41 i}{2 \cdot 8} : \frac{1}{4} + i \frac{41}{4}

x = \frac{- ( - 4 ) - 4 41 i}{2 \cdot 8} : \frac{1}{4} - i \frac{41}{4}

The solutions to the quadratic equation are:

x = \frac{1}{4} + i \frac{41}{4}, x = \frac{1}{4} - i \frac{41}{4}

(y + 5)^{2} = 5

3 x^{3} + 6 x^{2} - 9 x

3 x^{2} = 3 - 42 x

- \frac{4}{3 i}

What are the solutions to the equation 8x^2-4x+21=0 ?
The solutions to the equation 8x^2-4x+21=0 are x= 1/4+i(sqrt(41))/4 ,x= 1/4-i(sqrt(41))/4
Find the zeros of 8x^2-4x+21=0
The zeros of 8x^2-4x+21=0 are x= 1/4+i(sqrt(41))/4 ,x= 1/4-i(sqrt(41))/4