E=19h-(11h+27)h=8

Solution for E=19h-(11h+27)h=8 equation:

=19E-(11E+27)E=8

We move all terms to the left:

-(19E-(11E+27)E)=0


We calculate terms in parentheses: -(19E-(11E+27)E), so:

19E-(11E+27)E

We multiply parentheses

-11E^2+19E-27E

We add all the numbers together, and all the variables

-11E^2-8E

Back to the equation:

-(-11E^2-8E)


We get rid of parentheses

11E^2+8E=0

a = 11; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·11·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$E_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$E_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$E_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*11}=\frac{-16}{22}
=-8/11
$
$E_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*11}=\frac{0}{22}
=0
$