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w^2+10w+1=0
a = 1; b = 10; c = +1;
Δ = b2-4ac
Δ = 102-4·1·1
Δ = 96
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{6}}{2*1}=\frac{-10-4\sqrt{6}}{2} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{6}}{2*1}=\frac{-10+4\sqrt{6}}{2} $
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