What are the solutions to the equation 11=x^2+x-1 ?

The solutions to the equation 11=x^2+x-1 are x=3,x=-4

Find the zeros of 11=x^2+x-1

The zeros of 11=x^2+x-1 are x=3,x=-4

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11 = x^{2} + x - 1

x = 3, x = - 4

Solution steps

11 = x^{2} + x - 1

Switch sides

x^{2} + x - 1 = 11

Move

11

to the left side

x^{2} + x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 12) = 7

x_{1, 2} = \frac{- 1 \pm 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 7}{2 \cdot 1}, x_{2} = \frac{- 1 - 7}{2 \cdot 1}

x = \frac{- 1 + 7}{2 \cdot 1} : 3

x = \frac{- 1 - 7}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 3, x = - 4

5 x^{2} - 3 x = x^{2} - 4 x + 7

4 y^{2} + 12 y - 3 = 5 y

2 b^{2} - 32 = 0

20 x^{2} - 7 x = 14

x^{2} - 8 x - 3 = 6

What are the solutions to the equation 11=x^2+x-1 ?
The solutions to the equation 11=x^2+x-1 are x=3,x=-4
Find the zeros of 11=x^2+x-1
The zeros of 11=x^2+x-1 are x=3,x=-4