(6h-11)(h-+3)=0

Solution for (6h-11)(h-+3)=0 equation:

(6h-11)(h-+3)=0

We add all the numbers together, and all the variables

(6h-11)(+h)=0

We multiply parentheses ..

(+6h^2-11h)=0

We get rid of parentheses

6h^2-11h=0

a = 6; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·6·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*6}=\frac{0}{12}
=0
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*6}=\frac{22}{12}
=1+5/6
$