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(H)=-16H^2+120H
We move all terms to the left:
(H)-(-16H^2+120H)=0
We get rid of parentheses
16H^2-120H+H=0
We add all the numbers together, and all the variables
16H^2-119H=0
a = 16; b = -119; c = 0;
Δ = b2-4ac
Δ = -1192-4·16·0
Δ = 14161
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{14161}=119$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-119)-119}{2*16}=\frac{0}{32} =0 $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-119)+119}{2*16}=\frac{238}{32} =7+7/16 $
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