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

The solutions to the equation x^2-x=3 are x=(1+sqrt(13))/2 ,x=(1-sqrt(13))/2

Find the zeros of x^2-x=3

The zeros of x^2-x=3 are x=(1+sqrt(13))/2 ,x=(1-sqrt(13))/2

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x^{2} - x = 3

x = \frac{1 + 13}{2}, x = \frac{1 - 13}{2}

Solution steps

x^{2} - x = 3

Move

3

to the left side

x^{2} - x - 3 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 3) = 13

x_{1, 2} = \frac{- ( - 1 ) \pm 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 13}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 13}{2 \cdot 1}

x = \frac{- ( - 1 ) + 13}{2 \cdot 1} : \frac{1 + 13}{2}

x = \frac{- ( - 1 ) - 13}{2 \cdot 1} : \frac{1 - 13}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 13}{2}, x = \frac{1 - 13}{2}

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

24 x^{2} + 14 x y + 2 y^{2}

2 x^{2} = 18

x^{2} - 6 x - 2 = 0

6 a^{2} + 5 a - 6

What are the solutions to the equation x^2-x=3 ?
The solutions to the equation x^2-x=3 are x=(1+sqrt(13))/2 ,x=(1-sqrt(13))/2
Find the zeros of x^2-x=3
The zeros of x^2-x=3 are x=(1+sqrt(13))/2 ,x=(1-sqrt(13))/2