(X-3)2=x2+(-3)2

Solution for (X-3)2=x2+(-3)2 equation:

(X-3)2=X2+(-3)2

We move all terms to the left:

(X-3)2-(X2+(-3)2)=0

We multiply parentheses

2X-(X2+(-3)2)-6=0


We calculate terms in parentheses: -(X2+(-3)2), so:

X2+(-3)2

We add all the numbers together, and all the variables

X^2-6

Back to the equation:

-(X^2-6)


We get rid of parentheses

-X^2+2X+6-6=0

We add all the numbers together, and all the variables

-1X^2+2X=0

a = -1; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-1)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-1}=\frac{-4}{-2}
=+2
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-1}=\frac{0}{-2}
=0
$