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

The solutions to the equation 2b^2-3b+1=0 are b=1,b= 1/2

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

The zeros of 2b^2-3b+1=0 are b=1,b= 1/2

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2 b^{2} - 3 b + 1 = 0

b = 1, b = \frac{1}{2}

Decimal Notation

b = 1, b = 0.5

Solution steps

2 b^{2} - 3 b + 1 = 0

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 2 \cdot 1 = 1

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

Separate the solutions

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

b = \frac{- ( - 3 ) + 1}{2 \cdot 2} : 1

b = \frac{- ( - 3 ) - 1}{2 \cdot 2} : \frac{1}{2}

The solutions to the quadratic equation are:

b = 1, b = \frac{1}{2}

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

a^{2} x^{2} - 2 a x - 3 = 0

y (\frac{1 + y}{2 - k}) = a

\frac{( x + 2 ) ^{2}}{5} + \frac{( x + 2 ) ^{2}}{3} = \frac{16}{3}

0.000006 x^{2} - 0.02 x + 25 = 10 - 0.04 x

What are the solutions to the equation 2b^2-3b+1=0 ?
The solutions to the equation 2b^2-3b+1=0 are b=1,b= 1/2
Find the zeros of 2b^2-3b+1=0
The zeros of 2b^2-3b+1=0 are b=1,b= 1/2