What are the solutions to the equation 0=4x^2-3x+1 ?

The solutions to the equation 0=4x^2-3x+1 are x= 3/8+i(sqrt(7))/8 ,x= 3/8-i(sqrt(7))/8

Find the zeros of 0=4x^2-3x+1

The zeros of 0=4x^2-3x+1 are x= 3/8+i(sqrt(7))/8 ,x= 3/8-i(sqrt(7))/8

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0 = 4 x^{2} - 3 x + 1

x = \frac{3}{8} + i \frac{7}{8}, x = \frac{3}{8} - i \frac{7}{8}

Solution steps

0 = 4 x^{2} - 3 x + 1

Switch sides

4 x^{2} - 3 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

Simplify

(- 3)^{2} - 4 \cdot 4 \cdot 1 : 7 i

x_{1, 2} = \frac{- ( - 3 ) \pm 7 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 7 i}{2 \cdot 4}, x_{2} = \frac{- ( - 3 ) - 7 i}{2 \cdot 4}

x = \frac{- ( - 3 ) + 7 i}{2 \cdot 4} : \frac{3}{8} + i \frac{7}{8}

x = \frac{- ( - 3 ) - 7 i}{2 \cdot 4} : \frac{3}{8} - i \frac{7}{8}

The solutions to the quadratic equation are:

x = \frac{3}{8} + i \frac{7}{8}, x = \frac{3}{8} - i \frac{7}{8}

8 t^{2} + 24 t = 2 t - 5

3 y^{2} - 5 y = 0

5 x^{2} - 7 = - 4

x, (x - 1) (x + k) = - 4

x^{2} - \frac{1}{4} x = 0

What are the solutions to the equation 0=4x^2-3x+1 ?
The solutions to the equation 0=4x^2-3x+1 are x= 3/8+i(sqrt(7))/8 ,x= 3/8-i(sqrt(7))/8
Find the zeros of 0=4x^2-3x+1
The zeros of 0=4x^2-3x+1 are x= 3/8+i(sqrt(7))/8 ,x= 3/8-i(sqrt(7))/8