(5x-45)(5x-45)+22^2=(2x+3^2)+22^2

Solution for (5x-45)(5x-45)+22^2=(2x+3^2)+22^2 equation:

(5x-45)(5x-45)+22^2=(2x+3^2)+22^2

We move all terms to the left:

(5x-45)(5x-45)+22^2-((2x+3^2)+22^2)=0

We add all the numbers together, and all the variables

(5x-45)(5x-45)-((2x+3^2)+22^2)+484=0

We multiply parentheses ..

(+25x^2-225x-225x+2025)-((2x+3^2)+22^2)+484=0


We calculate terms in parentheses: -((2x+3^2)+22^2), so:

(2x+3^2)+22^2

We add all the numbers together, and all the variables

(2x+3^2)+484

We get rid of parentheses

2x+484+3^2

We add all the numbers together, and all the variables

2x+493

Back to the equation:

-(2x+493)


We get rid of parentheses

25x^2-225x-225x-2x+2025-493+484=0

We add all the numbers together, and all the variables

25x^2-452x+2016=0

a = 25; b = -452; c = +2016;
Δ = b2-4ac
Δ = -4522-4·25·2016
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-452)-52}{2*25}=\frac{400}{50}
=8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-452)+52}{2*25}=\frac{504}{50}
=10+2/25
$