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

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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 $

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