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(H)=200-16H^2
We move all terms to the left:
(H)-(200-16H^2)=0
We get rid of parentheses
16H^2+H-200=0
a = 16; b = 1; c = -200;
Δ = b2-4ac
Δ = 12-4·16·(-200)
Δ = 12801
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{12801}}{2*16}=\frac{-1-\sqrt{12801}}{32} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{12801}}{2*16}=\frac{-1+\sqrt{12801}}{32} $
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