x-1+x-2=x^2+(-3^2)

Solution for x-1+x-2=x^2+(-3^2) equation:

x-1+x-2=x^2+(-3^2)

We move all terms to the left:

x-1+x-2-(x^2+(-3^2))=0

We add all the numbers together, and all the variables

2x-(x^2+(-3^2))-3=0


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

x^2+(-3^2)

We add all the numbers together, and all the variables

x^2-9

Back to the equation:

-(x^2-9)


We get rid of parentheses

-x^2+2x+9-3=0

We add all the numbers together, and all the variables

-1x^2+2x+6=0

a = -1; b = 2; c = +6;
Δ = b2-4ac
Δ = 22-4·(-1)·6
Δ = 28
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{7}}{2*-1}=\frac{-2-2\sqrt{7}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{7}}{2*-1}=\frac{-2+2\sqrt{7}}{-2}
$