h^2+20h+98=(10+h)

Solution for h^2+20h+98=(10+h) equation:

h^2+20h+98=(10+h)

We move all terms to the left:

h^2+20h+98-((10+h))=0

We add all the numbers together, and all the variables

h^2+20h-((h+10))+98=0


We calculate terms in parentheses: -((h+10)), so:

(h+10)

We get rid of parentheses

h+10

Back to the equation:

-(h+10)


We get rid of parentheses

h^2+20h-h-10+98=0

We add all the numbers together, and all the variables

h^2+19h+88=0

a = 1; b = 19; c = +88;
Δ = b2-4ac
Δ = 192-4·1·88
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{-(19)-3}{2*1}=\frac{-22}{2}
=-11
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+3}{2*1}=\frac{-16}{2}
=-8
$