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=60H-16H^2
We move all terms to the left:
-(60H-16H^2)=0
We get rid of parentheses
16H^2-60H=0
a = 16; b = -60; c = 0;
Δ = b2-4ac
Δ = -602-4·16·0
Δ = 3600
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}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{3600}=60$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-60}{2*16}=\frac{0}{32} =0 $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+60}{2*16}=\frac{120}{32} =3+3/4 $
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