What are the solutions to the equation 2x^2-5x+1=3 ?

The solutions to the equation 2x^2-5x+1=3 are x=(5+sqrt(41))/4 ,x=(5-sqrt(41))/4

Find the zeros of 2x^2-5x+1=3

The zeros of 2x^2-5x+1=3 are x=(5+sqrt(41))/4 ,x=(5-sqrt(41))/4

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2 x^{2} - 5 x + 1 = 3

x = \frac{5 + 41}{4}, x = \frac{5 - 41}{4}

Decimal Notation

x = 2.85078 \dots, x = - 0.35078 \dots

Solution steps

2 x^{2} - 5 x + 1 = 3

Move

3

to the left side

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

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 2 (- 2) = 41

x_{1, 2} = \frac{- ( - 5 ) \pm 41}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 41}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 41}{2 \cdot 2}

x = \frac{- ( - 5 ) + 41}{2 \cdot 2} : \frac{5 + 41}{4}

x = \frac{- ( - 5 ) - 41}{2 \cdot 2} : \frac{5 - 41}{4}

The solutions to the quadratic equation are:

x = \frac{5 + 41}{4}, x = \frac{5 - 41}{4}

- 15 t - 6

v, m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I w^{2}

(27 x^{3})^{\frac{2}{3}}

5 x^{2} + 5 x y - 5 x^{2} y

What are the solutions to the equation 2x^2-5x+1=3 ?
The solutions to the equation 2x^2-5x+1=3 are x=(5+sqrt(41))/4 ,x=(5-sqrt(41))/4
Find the zeros of 2x^2-5x+1=3
The zeros of 2x^2-5x+1=3 are x=(5+sqrt(41))/4 ,x=(5-sqrt(41))/4