((10+5h)/(h^2+4h+5))-(10/5)=0

Solution for ((10+5h)/(h^2+4h+5))-(10/5)=0 equation:

((10+5h)/(h^2+4h+5))-(10/5)=0


Domain of the equation: (h^2+4h+5))!=0

h∈R


We add all the numbers together, and all the variables

((5h+10)/(h^2+4h+5))-2=0

We multiply all the terms by the denominator


((5h+10)-2*(h^2+4h+5))=0


We calculate terms in parentheses: +((5h+10)-2*(h^2+4h+5)), so:

(5h+10)-2*(h^2+4h+5)

We multiply parentheses

-2h^2+(5h+10)-8h-10

We get rid of parentheses

-2h^2+5h-8h+10-10

We add all the numbers together, and all the variables

-2h^2-3h

Back to the equation:

+(-2h^2-3h)


We get rid of parentheses

-2h^2-3h=0

a = -2; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·(-2)·0
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{-(-3)-3}{2*-2}=\frac{0}{-4}
=0
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*-2}=\frac{6}{-4}
=-1+1/2
$