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a2+13a+40=0
We add all the numbers together, and all the variables
a^2+13a+40=0
a = 1; b = 13; c = +40;
Δ = b2-4ac
Δ = 132-4·1·40
Δ = 9
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{9}=3$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-3}{2*1}=\frac{-16}{2} =-8 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+3}{2*1}=\frac{-10}{2} =-5 $
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