X+1=4(1-x)(3-x)

Solution for X+1=4(1-x)(3-x) equation:

X+1=4(1-X)(3-X)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


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

4(+X^2-3X-1X+3)

We multiply parentheses

4X^2-12X-4X+12

We add all the numbers together, and all the variables

4X^2-16X+12

Back to the equation:

-(4X^2-16X+12)


We add all the numbers together, and all the variables

X-(4X^2-16X+12)+1=0

We get rid of parentheses

-4X^2+X+16X-12+1=0

We add all the numbers together, and all the variables

-4X^2+17X-11=0

a = -4; b = 17; c = -11;
Δ = b2-4ac
Δ = 172-4·(-4)·(-11)
Δ = 113
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{-(17)-\sqrt{113}}{2*-4}=\frac{-17-\sqrt{113}}{-8}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{113}}{2*-4}=\frac{-17+\sqrt{113}}{-8}
$