((2x-10)*(2x-10))+x^2=625

Solution for ((2x-10)*(2x-10))+x^2=625 equation:

((2x-10)(2x-10))+x^2=625

We move all terms to the left:

((2x-10)(2x-10))+x^2-(625)=0

We multiply parentheses ..

x^2+((+4x^2-20x-20x+100))-625=0


We calculate terms in parentheses: +((+4x^2-20x-20x+100)), so:

(+4x^2-20x-20x+100)

We get rid of parentheses

4x^2-20x-20x+100

We add all the numbers together, and all the variables

4x^2-40x+100

Back to the equation:

+(4x^2-40x+100)


We get rid of parentheses

x^2+4x^2-40x+100-625=0

We add all the numbers together, and all the variables

5x^2-40x-525=0

a = 5; b = -40; c = -525;
Δ = b2-4ac
Δ = -402-4·5·(-525)
Δ = 12100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12100}=110$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-110}{2*5}=\frac{-70}{10}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+110}{2*5}=\frac{150}{10}
=15
$