4d+2=(d-1)(d-3)

Solution for 4d+2=(d-1)(d-3) equation:

4d+2=(d-1)(d-3)

We move all terms to the left:

4d+2-((d-1)(d-3))=0

We multiply parentheses ..

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


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

(+d^2-3d-1d+3)

We get rid of parentheses

d^2-3d-1d+3

We add all the numbers together, and all the variables

d^2-4d+3

Back to the equation:

-(d^2-4d+3)


We add all the numbers together, and all the variables

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

We get rid of parentheses

-d^2+4d+4d-3+2=0

We add all the numbers together, and all the variables

-1d^2+8d-1=0

a = -1; b = 8; c = -1;
Δ = b2-4ac
Δ = 82-4·(-1)·(-1)
Δ = 60
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{15}}{2*-1}=\frac{-8-2\sqrt{15}}{-2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{15}}{2*-1}=\frac{-8+2\sqrt{15}}{-2}
$