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=3H^2+3H
We move all terms to the left:
-(3H^2+3H)=0
We get rid of parentheses
-3H^2-3H=0
a = -3; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·(-3)·0
Δ = 9
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{9}=3$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*-3}=\frac{0}{-6} =0 $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*-3}=\frac{6}{-6} =-1 $
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