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-35+2w^2+11w=0
a = 2; b = 11; c = -35;
Δ = b2-4ac
Δ = 112-4·2·(-35)
Δ = 401
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{401}}{2*2}=\frac{-11-\sqrt{401}}{4} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{401}}{2*2}=\frac{-11+\sqrt{401}}{4} $
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