(h-6)(h+7)=0

Solution for (h-6)(h+7)=0 equation:

(h-6)(h+7)=0

We multiply parentheses ..

(+h^2+7h-6h-42)=0

We get rid of parentheses

h^2+7h-6h-42=0

We add all the numbers together, and all the variables

h^2+h-42=0

a = 1; b = 1; c = -42;
Δ = b2-4ac
Δ = 12-4·1·(-42)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-13}{2*1}=\frac{-14}{2}
=-7
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+13}{2*1}=\frac{12}{2}
=6
$