2(5-2h)=20-2h(h-1)

Solution for 2(5-2h)=20-2h(h-1) equation:

2(5-2h)=20-2h(h-1)

We move all terms to the left:

2(5-2h)-(20-2h(h-1))=0

We add all the numbers together, and all the variables

2(-2h+5)-(20-2h(h-1))=0

We multiply parentheses

-4h-(20-2h(h-1))+10=0


We calculate terms in parentheses: -(20-2h(h-1)), so:

20-2h(h-1)

determiningTheFunctionDomain
-2h(h-1)+20

We multiply parentheses

-2h^2+2h+20

Back to the equation:

-(-2h^2+2h+20)


We get rid of parentheses

2h^2-2h-4h-20+10=0

We add all the numbers together, and all the variables

2h^2-6h-10=0

a = 2; b = -6; c = -10;
Δ = b2-4ac
Δ = -62-4·2·(-10)
Δ = 116
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:

$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{29}}{2*2}=\frac{6-2\sqrt{29}}{4}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{29}}{2*2}=\frac{6+2\sqrt{29}}{4}
$