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(H)=-3H^2+6H
We move all terms to the left:
(H)-(-3H^2+6H)=0
We get rid of parentheses
3H^2-6H+H=0
We add all the numbers together, and all the variables
3H^2-5H=0
a = 3; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·3·0
Δ = 25
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{25}=5$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*3}=\frac{0}{6} =0 $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*3}=\frac{10}{6} =1+2/3 $
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