What are the solutions to the equation x^2+x=10 ?

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

Find the zeros of x^2+x=10

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

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x^{2} + x = 10

x = \frac{- 1 + 41}{2}, x = \frac{- 1 - 41}{2}

Decimal Notation

x = 2.70156 \dots, x = - 3.70156 \dots

Solution steps

x^{2} + x = 10

Move

10

to the left side

x^{2} + x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 10 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 10) = 41

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

Separate the solutions

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

x = \frac{- 1 + 41}{2 \cdot 1} : \frac{- 1 + 41}{2}

x = \frac{- 1 - 41}{2 \cdot 1} : \frac{- 1 - 41}{2}

The solutions to the quadratic equation are:

x = \frac{- 1 + 41}{2}, x = \frac{- 1 - 41}{2}

16 m^{2} - 9

12 X + 9 = 7 + 16 X

4^{2} \cdot 4^{5}

6 4^{- \frac{5}{6}}

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