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(H)=4H^2-36
We move all terms to the left:
(H)-(4H^2-36)=0
We get rid of parentheses
-4H^2+H+36=0
a = -4; b = 1; c = +36;
Δ = b2-4ac
Δ = 12-4·(-4)·36
Δ = 577
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{-(1)-\sqrt{577}}{2*-4}=\frac{-1-\sqrt{577}}{-8} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{577}}{2*-4}=\frac{-1+\sqrt{577}}{-8} $
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