What are the solutions to the equation 2x^2-5x+1=4 ?

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

Find the zeros of 2x^2-5x+1=4

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

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2 x^{2} - 5 x + 1 = 4

x = 3, x = - \frac{1}{2}

Decimal Notation

x = 3, x = - 0.5

Solution steps

2 x^{2} - 5 x + 1 = 4

Move

4

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

(- 5)^{2} - 4 \cdot 2 (- 3) = 7

x_{1, 2} = \frac{- ( - 5 ) \pm 7}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 7}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 7}{2 \cdot 2}

x = \frac{- ( - 5 ) + 7}{2 \cdot 2} : 3

x = \frac{- ( - 5 ) - 7}{2 \cdot 2} : - \frac{1}{2}

The solutions to the quadratic equation are:

x = 3, x = - \frac{1}{2}

8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1

a^{2} - 3 a + 2

16 - (2 a + b)^{2}

x^{2} + x = - 9 - 16

x + 9 = 3 - 5 x

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