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w^2+20w+30=0
a = 1; b = 20; c = +30;
Δ = b2-4ac
Δ = 202-4·1·30
Δ = 280
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{280}=\sqrt{4*70}=\sqrt{4}*\sqrt{70}=2\sqrt{70}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{70}}{2*1}=\frac{-20-2\sqrt{70}}{2} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{70}}{2*1}=\frac{-20+2\sqrt{70}}{2} $
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