What are the solutions to the equation roots 2x^2-3x+1 ?

The solutions to the equation roots 2x^2-3x+1 are x=1,x= 1/2

Find the zeros of roots 2x^2-3x+1

The zeros of roots 2x^2-3x+1 are x=1,x= 1/2

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roots

2 x^{2} - 3 x + 1

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

Solution steps

2 x^{2} - 3 x + 1

The roots are the intercepts with the x-axis

(y = 0)

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

x^{2} - 17 x + 40 = 0

x^{2} + 24 x + 140 = 0

r^{2} + 2 r^{2} = 10

x^{2} - 9 = 35

(3 z - 15)^{2} + 43 = - 2

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