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(H)=H2+6H-27
We move all terms to the left:
(H)-(H2+6H-27)=0
We add all the numbers together, and all the variables
-(+H^2+6H-27)+H=0
We get rid of parentheses
-H^2-6H+H+27=0
We add all the numbers together, and all the variables
-1H^2-5H+27=0
a = -1; b = -5; c = +27;
Δ = b2-4ac
Δ = -52-4·(-1)·27
Δ = 133
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}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{133}}{2*-1}=\frac{5-\sqrt{133}}{-2} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{133}}{2*-1}=\frac{5+\sqrt{133}}{-2} $
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