H5x(x-12)=260+4x

Solution for H5x(x-12)=260+4x equation:

5H(H-12)=260+4H

We move all terms to the left:

5H(H-12)-(260+4H)=0

We add all the numbers together, and all the variables

5H(H-12)-(4H+260)=0

We multiply parentheses

5H^2-60H-(4H+260)=0

We get rid of parentheses

5H^2-60H-4H-260=0

We add all the numbers together, and all the variables

5H^2-64H-260=0

a = 5; b = -64; c = -260;
Δ = b2-4ac
Δ = -642-4·5·(-260)
Δ = 9296
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:

$\sqrt{\Delta}=\sqrt{9296}=\sqrt{16*581}=\sqrt{16}*\sqrt{581}=4\sqrt{581}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-4\sqrt{581}}{2*5}=\frac{64-4\sqrt{581}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+4\sqrt{581}}{2*5}=\frac{64+4\sqrt{581}}{10}
$