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(H)=-16H^2+48H+28
We move all terms to the left:
(H)-(-16H^2+48H+28)=0
We get rid of parentheses
16H^2-48H+H-28=0
We add all the numbers together, and all the variables
16H^2-47H-28=0
a = 16; b = -47; c = -28;
Δ = b2-4ac
Δ = -472-4·16·(-28)
Δ = 4001
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{-(-47)-\sqrt{4001}}{2*16}=\frac{47-\sqrt{4001}}{32} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+\sqrt{4001}}{2*16}=\frac{47+\sqrt{4001}}{32} $
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