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(H)=-16H^2-30H+60
We move all terms to the left:
(H)-(-16H^2-30H+60)=0
We get rid of parentheses
16H^2+30H+H-60=0
We add all the numbers together, and all the variables
16H^2+31H-60=0
a = 16; b = 31; c = -60;
Δ = b2-4ac
Δ = 312-4·16·(-60)
Δ = 4801
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{-(31)-\sqrt{4801}}{2*16}=\frac{-31-\sqrt{4801}}{32} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+\sqrt{4801}}{2*16}=\frac{-31+\sqrt{4801}}{32} $
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