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2w^2-3w-50=0
a = 2; b = -3; c = -50;
Δ = b2-4ac
Δ = -32-4·2·(-50)
Δ = 409
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{-(-3)-\sqrt{409}}{2*2}=\frac{3-\sqrt{409}}{4} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{409}}{2*2}=\frac{3+\sqrt{409}}{4} $
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