Answer :

Answer:

13 and -14 satisfy this condition

Step-by-step explanation:

Let's represent that number as x

and the square of x is x^2

So,

x + x^2 = 182

Subtract 182 from both sides

x + x^2 - 182 = 182 - 182

x + x^2 - 182 = 0

rearrange the quadratic equation

x^2 + x -182 = 0

let's use the quadratic formula

[tex]\frac{-b+\sqrt{b^2-4ac} }{2a}[/tex] or [tex]\frac{-b-\sqrt{b^2-4ac} }{2a}[/tex]

a = 1, b = 1, c = -182

[tex]\frac{-1+\sqrt{1^2-4*1*(-182)} }{2*1}[/tex] or [tex]\frac{-1-\sqrt{1^2-4*1*(-182)} }{2*1}[/tex]

[tex]\frac{-1+\sqrt{1+728} }{2}[/tex] or [tex]\frac{-1-\sqrt{1+728} }{2}[/tex]

[tex]\frac{-1+\sqrt{729} }{2}[/tex] or [tex]\frac{-1-\sqrt{729} }{2}[/tex]

[tex]\frac{-1+{27} }{2}[/tex] or [tex]\frac{-1-{27} }{2}[/tex]

[tex]\frac{26}{2}[/tex] or [tex]\frac{-28}{2}[/tex]

13 or - 14

Lets check

13 + 13^2 = 13 + 169

= 182

Also,

-14 + (-14^2) = -14 + 196

= 182

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Rewritten by : Barada