Let
be the unknown number. So, three times that number means
, and the square of the number is ![x^2](https://tex.z-dn.net/?f=%20x%5E2%20)
We have to sum 528 and three times the number, so we have ![528+3x](https://tex.z-dn.net/?f=%20528%2B3x%20)
Then, we have to subtract this number from
, so we have
![x^2-(3x+528)](https://tex.z-dn.net/?f=%20x%5E2-%283x%2B528%29%20)
The result is 120, so the equation is
![x^2 - 3x - 528 = 120 \iff x^2 - 3x - 648 = 0](https://tex.z-dn.net/?f=%20x%5E2%20-%203x%20-%20528%20%3D%20120%20%5Ciff%20x%5E2%20-%203x%20-%20648%20%3D%200%20)
This is a quadratic equation, i.e. an equation like
. These equation can be solved - assuming they have a solution - with the following formula
![x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=%20x_%7B1%2C2%7D%20%3D%20%5Cdfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D%20)
If you plug the values from your equation, you have
![x_{1,2} = \dfrac{3\pm\sqrt{9-4\cdot(-648)}}{2} = \dfrac{3\pm\sqrt{9+2592}}{2} = \dfrac{3\pm\sqrt{2601}}{2} = \dfrac{3\pm51}{2}](https://tex.z-dn.net/?f=x_%7B1%2C2%7D%20%3D%20%5Cdfrac%7B3%5Cpm%5Csqrt%7B9-4%5Ccdot%28-648%29%7D%7D%7B2%7D%20%3D%20%5Cdfrac%7B3%5Cpm%5Csqrt%7B9%2B2592%7D%7D%7B2%7D%20%3D%20%5Cdfrac%7B3%5Cpm%5Csqrt%7B2601%7D%7D%7B2%7D%20%3D%20%5Cdfrac%7B3%5Cpm51%7D%7B2%7D)
So, the two solutions would be
![x = \dfrac{3+51}{2} = \dfrac{54}{2} = 27](https://tex.z-dn.net/?f=%20x%20%3D%20%5Cdfrac%7B3%2B51%7D%7B2%7D%20%3D%20%5Cdfrac%7B54%7D%7B2%7D%20%3D%2027%20)
![x = \dfrac{3-51}{2} = \dfrac{-48}{2} = -24](https://tex.z-dn.net/?f=%20x%20%3D%20%5Cdfrac%7B3-51%7D%7B2%7D%20%3D%20%5Cdfrac%7B-48%7D%7B2%7D%20%3D%20-24%20)
But we know that x is positive, so we only accept the solution ![x = 27](https://tex.z-dn.net/?f=%20x%20%3D%2027%20)