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2w^2-5-9w=0
a = 2; b = -9; c = -5;
Δ = b2-4ac
Δ = -92-4·2·(-5)
Δ = 121
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{121}=11$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-11}{2*2}=\frac{-2}{4} =-1/2 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+11}{2*2}=\frac{20}{4} =5 $
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