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

The solutions to the equation 5x^2-x+1=3 are x=(1+sqrt(41))/(10),x=(1-sqrt(41))/(10)

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

The zeros of 5x^2-x+1=3 are x=(1+sqrt(41))/(10),x=(1-sqrt(41))/(10)

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

x = \frac{1 + 41}{10}, x = \frac{1 - 41}{10}

Decimal Notation

x = 0.74031 \dots, x = - 0.54031 \dots

Solution steps

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

Move

3

to the left side

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 5 (- 2) = 41

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

Separate the solutions

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

x = \frac{- ( - 1 ) + 41}{2 \cdot 5} : \frac{1 + 41}{10}

x = \frac{- ( - 1 ) - 41}{2 \cdot 5} : \frac{1 - 41}{10}

The solutions to the quadratic equation are:

x = \frac{1 + 41}{10}, x = \frac{1 - 41}{10}

3 (x - 3) = 11 - 5 (3 x + 1)

4 p^{2} = - 17 p - 4

∣ x + 2 ∣ = 3

\frac{- 2}{5 ^{2}}

\frac{2 x - 2}{3} = ∣ - 4 ∣

What are the solutions to the equation 5x^2-x+1=3 ?
The solutions to the equation 5x^2-x+1=3 are x=(1+sqrt(41))/(10),x=(1-sqrt(41))/(10)
Find the zeros of 5x^2-x+1=3
The zeros of 5x^2-x+1=3 are x=(1+sqrt(41))/(10),x=(1-sqrt(41))/(10)