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(H)=-16H^2+12H+30
We move all terms to the left:
(H)-(-16H^2+12H+30)=0
We get rid of parentheses
16H^2-12H+H-30=0
We add all the numbers together, and all the variables
16H^2-11H-30=0
a = 16; b = -11; c = -30;
Δ = b2-4ac
Δ = -112-4·16·(-30)
Δ = 2041
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{-(-11)-\sqrt{2041}}{2*16}=\frac{11-\sqrt{2041}}{32} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{2041}}{2*16}=\frac{11+\sqrt{2041}}{32} $
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