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(H)=-16H^2+6H+10
We move all terms to the left:
(H)-(-16H^2+6H+10)=0
We get rid of parentheses
16H^2-6H+H-10=0
We add all the numbers together, and all the variables
16H^2-5H-10=0
a = 16; b = -5; c = -10;
Δ = b2-4ac
Δ = -52-4·16·(-10)
Δ = 665
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{-(-5)-\sqrt{665}}{2*16}=\frac{5-\sqrt{665}}{32} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{665}}{2*16}=\frac{5+\sqrt{665}}{32} $
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