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50=2h^2+10h
We move all terms to the left:
50-(2h^2+10h)=0
We get rid of parentheses
-2h^2-10h+50=0
a = -2; b = -10; c = +50;
Δ = b2-4ac
Δ = -102-4·(-2)·50
Δ = 500
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}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10\sqrt{5}}{2*-2}=\frac{10-10\sqrt{5}}{-4} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10\sqrt{5}}{2*-2}=\frac{10+10\sqrt{5}}{-4} $
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