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(H)=-16H^2+1020
We move all terms to the left:
(H)-(-16H^2+1020)=0
We get rid of parentheses
16H^2+H-1020=0
a = 16; b = 1; c = -1020;
Δ = b2-4ac
Δ = 12-4·16·(-1020)
Δ = 65281
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{65281}}{2*16}=\frac{-1-\sqrt{65281}}{32} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{65281}}{2*16}=\frac{-1+\sqrt{65281}}{32} $
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