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(H)=-1/5H-7
We move all terms to the left:
(H)-(-1/5H-7)=0
Domain of the equation: 5H-7)!=0We get rid of parentheses
H∈R
H+1/5H+7=0
We multiply all the terms by the denominator
H*5H+7*5H+1=0
Wy multiply elements
5H^2+35H+1=0
a = 5; b = 35; c = +1;
Δ = b2-4ac
Δ = 352-4·5·1
Δ = 1205
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{-(35)-\sqrt{1205}}{2*5}=\frac{-35-\sqrt{1205}}{10} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+\sqrt{1205}}{2*5}=\frac{-35+\sqrt{1205}}{10} $
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