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2w^2-2450=0
a = 2; b = 0; c = -2450;
Δ = b2-4ac
Δ = 02-4·2·(-2450)
Δ = 19600
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{19600}=140$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-140}{2*2}=\frac{-140}{4} =-35 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+140}{2*2}=\frac{140}{4} =35 $
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