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=16H^2+50H
We move all terms to the left:
-(16H^2+50H)=0
We get rid of parentheses
-16H^2-50H=0
a = -16; b = -50; c = 0;
Δ = b2-4ac
Δ = -502-4·(-16)·0
Δ = 2500
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{2500}=50$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-50}{2*-16}=\frac{0}{-32} =0 $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+50}{2*-16}=\frac{100}{-32} =-3+1/8 $
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