H(x-4)=(x-4)(x-4)-2(x-4)

Solution for H(x-4)=(x-4)(x-4)-2(x-4) equation:

(H-4)=(H-4)(H-4)-2(H-4)

We move all terms to the left:

(H-4)-((H-4)(H-4)-2(H-4))=0

We get rid of parentheses

H-((H-4)(H-4)-2(H-4))-4=0

We multiply parentheses ..

-((+H^2-4H-4H+16)-2(H-4))+H-4=0


We calculate terms in parentheses: -((+H^2-4H-4H+16)-2(H-4)), so:

(+H^2-4H-4H+16)-2(H-4)

We multiply parentheses

(+H^2-4H-4H+16)-2H+8

We get rid of parentheses

H^2-4H-4H-2H+16+8

We add all the numbers together, and all the variables

H^2-10H+24

Back to the equation:

-(H^2-10H+24)


We add all the numbers together, and all the variables

H-(H^2-10H+24)-4=0

We get rid of parentheses

-H^2+H+10H-24-4=0

We add all the numbers together, and all the variables

-1H^2+11H-28=0

a = -1; b = 11; c = -28;
Δ = b2-4ac
Δ = 112-4·(-1)·(-28)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-3}{2*-1}=\frac{-14}{-2}
=+7
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+3}{2*-1}=\frac{-8}{-2}
=+4
$