What are the solutions to the equation 0=8x^2-6x+4 ?

The solutions to the equation 0=8x^2-6x+4 are x= 3/8+i(sqrt(23))/8 ,x= 3/8-i(sqrt(23))/8

Find the zeros of 0=8x^2-6x+4

The zeros of 0=8x^2-6x+4 are x= 3/8+i(sqrt(23))/8 ,x= 3/8-i(sqrt(23))/8

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0 = 8 x^{2} - 6 x + 4

x = \frac{3}{8} + i \frac{23}{8}, x = \frac{3}{8} - i \frac{23}{8}

Solution steps

0 = 8 x^{2} - 6 x + 4

Switch sides

8 x^{2} - 6 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 8 \cdot 4}{2 \cdot 8}

Simplify

(- 6)^{2} - 4 \cdot 8 \cdot 4 : 223 i

x_{1, 2} = \frac{- ( - 6 ) \pm 2 23 i}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 2 23 i}{2 \cdot 8}, x_{2} = \frac{- ( - 6 ) - 2 23 i}{2 \cdot 8}

x = \frac{- ( - 6 ) + 2 23 i}{2 \cdot 8} : \frac{3}{8} + i \frac{23}{8}

x = \frac{- ( - 6 ) - 2 23 i}{2 \cdot 8} : \frac{3}{8} - i \frac{23}{8}

The solutions to the quadratic equation are:

x = \frac{3}{8} + i \frac{23}{8}, x = \frac{3}{8} - i \frac{23}{8}

\frac{x ^{2} - 4 x + 5}{x ^{2} - 3 x - 4} \leq 0

u^{2} + 9 u + 18

5 x + 2 > x - 6

3 x - 6 \leq - 3

(x + 3)^{2} + (x - 1)^{2} = 40

What are the solutions to the equation 0=8x^2-6x+4 ?
The solutions to the equation 0=8x^2-6x+4 are x= 3/8+i(sqrt(23))/8 ,x= 3/8-i(sqrt(23))/8
Find the zeros of 0=8x^2-6x+4
The zeros of 0=8x^2-6x+4 are x= 3/8+i(sqrt(23))/8 ,x= 3/8-i(sqrt(23))/8