(8h-1)-(h+3);h=3

Solution for (8h-1)-(h+3);h=3 equation:

(8h-1)-(h+3)h=3

We move all terms to the left:

(8h-1)-(h+3)h-(3)=0

We multiply parentheses

-h^2+(8h-1)-3h-3=0

We get rid of parentheses

-h^2+8h-3h-1-3=0

We add all the numbers together, and all the variables

-1h^2+5h-4=0

a = -1; b = 5; c = -4;
Δ = b2-4ac
Δ = 52-4·(-1)·(-4)
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{-(5)-3}{2*-1}=\frac{-8}{-2}
=+4
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3}{2*-1}=\frac{-2}{-2}
=1
$