Y=3-4x+x2

Solution for Y=3-4x+x2 equation:

=3-4Y+Y2

We move all terms to the left:

-(3-4Y+Y2)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

-Y^2+4Y-3=0

We add all the numbers together, and all the variables

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

a = -1; b = 4; c = -3;
Δ = b2-4ac
Δ = 42-4·(-1)·(-3)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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