h(h+6)=17+2h

Solution for h(h+6)=17+2h equation:

h(h+6)=17+2h

We move all terms to the left:

h(h+6)-(17+2h)=0

We add all the numbers together, and all the variables

h(h+6)-(2h+17)=0

We multiply parentheses

h^2+6h-(2h+17)=0

We get rid of parentheses

h^2+6h-2h-17=0

We add all the numbers together, and all the variables

h^2+4h-17=0

a = 1; b = 4; c = -17;
Δ = b2-4ac
Δ = 42-4·1·(-17)
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{21}}{2*1}=\frac{-4-2\sqrt{21}}{2}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{21}}{2*1}=\frac{-4+2\sqrt{21}}{2}
$