What are the solutions to the equation x^2=13+x ?

The solutions to the equation x^2=13+x are x=(1+sqrt(53))/2 ,x=(1-sqrt(53))/2

Find the zeros of x^2=13+x

The zeros of x^2=13+x are x=(1+sqrt(53))/2 ,x=(1-sqrt(53))/2

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x^{2} = 13 + x

x = \frac{1 + 53}{2}, x = \frac{1 - 53}{2}

Decimal Notation

x = 4.14005 \dots, x = - 3.14005 \dots

Solution steps

x^{2} = 13 + x

Move

x

to the left side

x^{2} - x = 13

Move

13

to the left side

x^{2} - x - 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 13 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 13) = 53

x_{1, 2} = \frac{- ( - 1 ) \pm 53}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 53}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 53}{2 \cdot 1}

x = \frac{- ( - 1 ) + 53}{2 \cdot 1} : \frac{1 + 53}{2}

x = \frac{- ( - 1 ) - 53}{2 \cdot 1} : \frac{1 - 53}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 53}{2}, x = \frac{1 - 53}{2}

10 = \frac{3}{7} (4 x + 35)

7 x \leq 5 x + 7 < 6 x + 8

1176

4 9^{- \frac{3}{2}}

\frac{2 ^{- 1}}{2 ^{- 2}}

What are the solutions to the equation x^2=13+x ?
The solutions to the equation x^2=13+x are x=(1+sqrt(53))/2 ,x=(1-sqrt(53))/2
Find the zeros of x^2=13+x
The zeros of x^2=13+x are x=(1+sqrt(53))/2 ,x=(1-sqrt(53))/2