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21a^2+10a+1=0
a = 21; b = 10; c = +1;
Δ = b2-4ac
Δ = 102-4·21·1
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4}{2*21}=\frac{-14}{42} =-1/3 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4}{2*21}=\frac{-6}{42} =-1/7 $
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