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w^2-3w-48=0
a = 1; b = -3; c = -48;
Δ = b2-4ac
Δ = -32-4·1·(-48)
Δ = 201
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{201}}{2*1}=\frac{3-\sqrt{201}}{2} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{201}}{2*1}=\frac{3+\sqrt{201}}{2} $
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