(2h-4)(2h-4)=18

Solution for (2h-4)(2h-4)=18 equation:

(2h-4)(2h-4)=18

We move all terms to the left:

(2h-4)(2h-4)-(18)=0

We multiply parentheses ..

(+4h^2-8h-8h+16)-18=0

We get rid of parentheses

4h^2-8h-8h+16-18=0

We add all the numbers together, and all the variables

4h^2-16h-2=0

a = 4; b = -16; c = -2;
Δ = b2-4ac
Δ = -162-4·4·(-2)
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-12\sqrt{2}}{2*4}=\frac{16-12\sqrt{2}}{8}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+12\sqrt{2}}{2*4}=\frac{16+12\sqrt{2}}{8}
$