1/2h+10=h+4

Solution for 1/2h+10=h+4 equation:

1/2h+10=h+4

We move all terms to the left:

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


Domain of the equation: 2h!=0

h!=0/2

h!=0

h∈R


We get rid of parentheses

1/2h-h-4+10=0

We multiply all the terms by the denominator


-h*2h-4*2h+10*2h+1=0

Wy multiply elements

-2h^2-8h+20h+1=0

We add all the numbers together, and all the variables

-2h^2+12h+1=0

a = -2; b = 12; c = +1;
Δ = b2-4ac
Δ = 122-4·(-2)·1
Δ = 152
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:

$\sqrt{\Delta}=\sqrt{152}=\sqrt{4*38}=\sqrt{4}*\sqrt{38}=2\sqrt{38}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{38}}{2*-2}=\frac{-12-2\sqrt{38}}{-4}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{38}}{2*-2}=\frac{-12+2\sqrt{38}}{-4}
$