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(H)=-16H^2+30H+6H
We move all terms to the left:
(H)-(-16H^2+30H+6H)=0
We get rid of parentheses
16H^2-30H-6H+H=0
We add all the numbers together, and all the variables
16H^2-35H=0
a = 16; b = -35; c = 0;
Δ = b2-4ac
Δ = -352-4·16·0
Δ = 1225
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{1225}=35$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-35}{2*16}=\frac{0}{32} =0 $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+35}{2*16}=\frac{70}{32} =2+3/16 $
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