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(H)=-16H^2+130
We move all terms to the left:
(H)-(-16H^2+130)=0
We get rid of parentheses
16H^2+H-130=0
a = 16; b = 1; c = -130;
Δ = b2-4ac
Δ = 12-4·16·(-130)
Δ = 8321
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{8321}}{2*16}=\frac{-1-\sqrt{8321}}{32} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{8321}}{2*16}=\frac{-1+\sqrt{8321}}{32} $
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