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5w^2-405=0
a = 5; b = 0; c = -405;
Δ = b2-4ac
Δ = 02-4·5·(-405)
Δ = 8100
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{8100}=90$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-90}{2*5}=\frac{-90}{10} =-9 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+90}{2*5}=\frac{90}{10} =9 $
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