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r2-13r+40=0
We add all the numbers together, and all the variables
r^2-13r+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:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{9}=3$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-3}{2*1}=\frac{10}{2} =5 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+3}{2*1}=\frac{16}{2} =8 $
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