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18w^2-50=0
a = 18; b = 0; c = -50;
Δ = b2-4ac
Δ = 02-4·18·(-50)
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{3600}=60$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-60}{2*18}=\frac{-60}{36} =-1+2/3 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+60}{2*18}=\frac{60}{36} =1+2/3 $
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