-x^2+4x-3=x^2-2x-3

Solution for -x^2+4x-3=x^2-2x-3 equation:

-x^2+4x-3=x^2-2x-3

We move all terms to the left:

-x^2+4x-3-(x^2-2x-3)=0

We add all the numbers together, and all the variables

-1x^2+4x-(x^2-2x-3)-3=0

We get rid of parentheses

-1x^2-x^2+4x+2x+3-3=0

We add all the numbers together, and all the variables

-2x^2+6x=0

a = -2; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-2)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-2}=\frac{-12}{-4}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-2}=\frac{0}{-4}
=0
$