(x^2=2x-5)-(3x+2)=x^2-1

Solution for (x^2=2x-5)-(3x+2)=x^2-1 equation:

(x^2=2x-5)-(3x+2)=x^2-1

We move all terms to the left:

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


We calculate terms in parentheses: +(x^2-(2x-5)-(3x+2)), so:

x^2-(2x-5)-(3x+2)

We get rid of parentheses

x^2-2x-3x+5-2

We add all the numbers together, and all the variables

x^2-5x+3

Back to the equation:

+(x^2-5x+3)


We get rid of parentheses

x^2-5x+3=0

a = 1; b = -5; c = +3;
Δ = b2-4ac
Δ = -52-4·1·3
Δ = 13
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{13}}{2*1}=\frac{5-\sqrt{13}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{13}}{2*1}=\frac{5+\sqrt{13}}{2}
$