Y=(4-x)(3+x)

Solution for Y=(4-x)(3+x) equation:

=(4-Y)(3+Y)

We move all terms to the left:

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

We add all the numbers together, and all the variables

-((-1Y+4)(Y+3))=0

We multiply parentheses ..

-((-1Y^2-3Y+4Y+12))=0


We calculate terms in parentheses: -((-1Y^2-3Y+4Y+12)), so:

(-1Y^2-3Y+4Y+12)

We get rid of parentheses

-1Y^2-3Y+4Y+12

We add all the numbers together, and all the variables

-1Y^2+Y+12

Back to the equation:

-(-1Y^2+Y+12)


We get rid of parentheses

1Y^2-Y-12=0

We add all the numbers together, and all the variables

Y^2-1Y-12=0

a = 1; b = -1; c = -12;
Δ = b2-4ac
Δ = -12-4·1·(-12)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-7}{2*1}=\frac{-6}{2}
=-3
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+7}{2*1}=\frac{8}{2}
=4
$