-x^2+4x-1=x^2-4x+5

Solution for -x^2+4x-1=x^2-4x+5 equation:

-x^2+4x-1=x^2-4x+5

We move all terms to the left:

-x^2+4x-1-(x^2-4x+5)=0

We add all the numbers together, and all the variables

-1x^2+4x-(x^2-4x+5)-1=0

We get rid of parentheses

-1x^2-x^2+4x+4x-5-1=0

We add all the numbers together, and all the variables

-2x^2+8x-6=0

a = -2; b = 8; c = -6;
Δ = b2-4ac
Δ = 82-4·(-2)·(-6)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4}{2*-2}=\frac{-12}{-4}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*-2}=\frac{-4}{-4}
=1
$