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a^2+14a+33=0
a = 1; b = 14; c = +33;
Δ = b2-4ac
Δ = 142-4·1·33
Δ = 64
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{64}=8$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-8}{2*1}=\frac{-22}{2} =-11 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+8}{2*1}=\frac{-6}{2} =-3 $
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