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w^2+3w-208=0
a = 1; b = 3; c = -208;
Δ = b2-4ac
Δ = 32-4·1·(-208)
Δ = 841
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{841}=29$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-29}{2*1}=\frac{-32}{2} =-16 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+29}{2*1}=\frac{26}{2} =13 $
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