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(H)=-16H^2+240H
We move all terms to the left:
(H)-(-16H^2+240H)=0
We get rid of parentheses
16H^2-240H+H=0
We add all the numbers together, and all the variables
16H^2-239H=0
a = 16; b = -239; c = 0;
Δ = b2-4ac
Δ = -2392-4·16·0
Δ = 57121
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{57121}=239$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-239)-239}{2*16}=\frac{0}{32} =0 $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-239)+239}{2*16}=\frac{478}{32} =14+15/16 $
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