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10w^2+27w+5=0
a = 10; b = 27; c = +5;
Δ = b2-4ac
Δ = 272-4·10·5
Δ = 529
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{529}=23$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-23}{2*10}=\frac{-50}{20} =-2+1/2 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+23}{2*10}=\frac{-4}{20} =-1/5 $
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