What are the solutions to the equation 2x^2-3x+1= 3/8 ?

The solutions to the equation 2x^2-3x+1= 3/8 are x= 5/4 ,x= 1/4

Find the zeros of 2x^2-3x+1= 3/8

The zeros of 2x^2-3x+1= 3/8 are x= 5/4 ,x= 1/4

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2 x^{2} - 3 x + 1 = \frac{3}{8}

x = \frac{5}{4}, x = \frac{1}{4}

Decimal Notation

x = 1.25, x = 0.25

Solution steps

2 x^{2} - 3 x + 1 = \frac{3}{8}

Multiply both sides by

8

16 x^{2} - 24 x + 8 = 3

Move

3

to the left side

16 x^{2} - 24 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 16 \cdot 5}{2 \cdot 16}

(- 24)^{2} - 4 \cdot 16 \cdot 5 = 16

x_{1, 2} = \frac{- ( - 24 ) \pm 16}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- ( - 24 ) + 16}{2 \cdot 16}, x_{2} = \frac{- ( - 24 ) - 16}{2 \cdot 16}

x = \frac{- ( - 24 ) + 16}{2 \cdot 16} : \frac{5}{4}

x = \frac{- ( - 24 ) - 16}{2 \cdot 16} : \frac{1}{4}

The solutions to the quadratic equation are:

x = \frac{5}{4}, x = \frac{1}{4}

(x - 2)^{2} + (x + 8)^{2} = 1 0^{2}

3 x^{2} + 3 x = 25 x - x^{2}

5 x^{2} + 5 = - 15

2 x^{2} - 6 = 3 x

5 x^{2} + x = - 5

What are the solutions to the equation 2x^2-3x+1= 3/8 ?
The solutions to the equation 2x^2-3x+1= 3/8 are x= 5/4 ,x= 1/4
Find the zeros of 2x^2-3x+1= 3/8
The zeros of 2x^2-3x+1= 3/8 are x= 5/4 ,x= 1/4