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(H)=0.00012H^2+200
We move all terms to the left:
(H)-(0.00012H^2+200)=0
We get rid of parentheses
-0.00012H^2+H-200=0
a = -0.00012; b = 1; c = -200;
Δ = b2-4ac
Δ = 12-4·(-0.00012)·(-200)
Δ = 0.904
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{0.904}}{2*-0.00012}=\frac{-1-\sqrt{0.904}}{-0.00024} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{0.904}}{2*-0.00012}=\frac{-1+\sqrt{0.904}}{-0.00024} $
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