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w^2+8w-20.25=0
a = 1; b = 8; c = -20.25;
Δ = b2-4ac
Δ = 82-4·1·(-20.25)
Δ = 145
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{-(8)-\sqrt{145}}{2*1}=\frac{-8-\sqrt{145}}{2} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+\sqrt{145}}{2*1}=\frac{-8+\sqrt{145}}{2} $
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