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a^2-13a-168=0
a = 1; b = -13; c = -168;
Δ = b2-4ac
Δ = -132-4·1·(-168)
Δ = 841
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{841}=29$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-29}{2*1}=\frac{-16}{2} =-8 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+29}{2*1}=\frac{42}{2} =21 $
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