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(5)=-4.9H^2+49H+100
We move all terms to the left:
(5)-(-4.9H^2+49H+100)=0
We get rid of parentheses
4.9H^2-49H-100+5=0
We add all the numbers together, and all the variables
4.9H^2-49H-95=0
a = 4.9; b = -49; c = -95;
Δ = b2-4ac
Δ = -492-4·4.9·(-95)
Δ = 4263
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}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4263}=\sqrt{49*87}=\sqrt{49}*\sqrt{87}=7\sqrt{87}$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-49)-7\sqrt{87}}{2*4.9}=\frac{49-7\sqrt{87}}{9.8} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+7\sqrt{87}}{2*4.9}=\frac{49+7\sqrt{87}}{9.8} $
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