124=(40-4h)(32-4h)

Solution for 124=(40-4h)(32-4h) equation:

124=(40-4h)(32-4h)

We move all terms to the left:

124-((40-4h)(32-4h))=0

We add all the numbers together, and all the variables

-((-4h+40)(-4h+32))+124=0

We multiply parentheses ..

-((+16h^2-128h-160h+1280))+124=0


We calculate terms in parentheses: -((+16h^2-128h-160h+1280)), so:

(+16h^2-128h-160h+1280)

We get rid of parentheses

16h^2-128h-160h+1280

We add all the numbers together, and all the variables

16h^2-288h+1280

Back to the equation:

-(16h^2-288h+1280)


We get rid of parentheses

-16h^2+288h-1280+124=0

We add all the numbers together, and all the variables

-16h^2+288h-1156=0

a = -16; b = 288; c = -1156;
Δ = b2-4ac
Δ = 2882-4·(-16)·(-1156)
Δ = 8960
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8960}=\sqrt{256*35}=\sqrt{256}*\sqrt{35}=16\sqrt{35}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(288)-16\sqrt{35}}{2*-16}=\frac{-288-16\sqrt{35}}{-32}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(288)+16\sqrt{35}}{2*-16}=\frac{-288+16\sqrt{35}}{-32}
$