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4w^2-20=0
a = 4; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·4·(-20)
Δ = 320
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{5}}{2*4}=\frac{0-8\sqrt{5}}{8} =-\frac{8\sqrt{5}}{8} =-\sqrt{5} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{5}}{2*4}=\frac{0+8\sqrt{5}}{8} =\frac{8\sqrt{5}}{8} =\sqrt{5} $
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