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

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

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

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

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4 b^{2} - 3 b + 2 = 0

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

Solution steps

4 b^{2} - 3 b + 2 = 0

Solve with the quadratic formula

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

Simplify

(- 3)^{2} - 4 \cdot 4 \cdot 2 : 23 i

b_{1, 2} = \frac{- ( - 3 ) \pm 23 i}{2 \cdot 4}

Separate the solutions

b_{1} = \frac{- ( - 3 ) + 23 i}{2 \cdot 4}, b_{2} = \frac{- ( - 3 ) - 23 i}{2 \cdot 4}

b = \frac{- ( - 3 ) + 23 i}{2 \cdot 4} : \frac{3}{8} + i \frac{23}{8}

b = \frac{- ( - 3 ) - 23 i}{2 \cdot 4} : \frac{3}{8} - i \frac{23}{8}

The solutions to the quadratic equation are:

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

x, (2 x + 2 y) (x - y) = (4 x y) (- x y)

- 2 x^{2} - 3 x + 5 = 0

2 (3 x + 5)^{2} - 44 = 16

x^{2} + ab = (a + b) x

6 x^{2} + 9 x - 6 = 0

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