What are the solutions to the equation x+x^2=182 ?

The solutions to the equation x+x^2=182 are x=13,x=-14

Find the zeros of x+x^2=182

The zeros of x+x^2=182 are x=13,x=-14

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x + x^{2} = 182

x = 13, x = - 14

Solution steps

x + x^{2} = 182

Move

182

to the left side

x + x^{2} - 182 = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} + x - 182 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 182 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 182) = 27

x_{1, 2} = \frac{- 1 \pm 27}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 27}{2 \cdot 1}, x_{2} = \frac{- 1 - 27}{2 \cdot 1}

x = \frac{- 1 + 27}{2 \cdot 1} : 13

x = \frac{- 1 - 27}{2 \cdot 1} : - 14

The solutions to the quadratic equation are:

x = 13, x = - 14

- 5 (3 n + 5) = 20

\frac{f}{2} - 1 \geq 2

25 y^{2} = - 16

x - 8 = 10

(c^{3})^{5} (d^{3})^{0}

What are the solutions to the equation x+x^2=182 ?
The solutions to the equation x+x^2=182 are x=13,x=-14
Find the zeros of x+x^2=182
The zeros of x+x^2=182 are x=13,x=-14