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r2-13r+30=0
We add all the numbers together, and all the variables
r^2-13r+30=0
a = 1; b = -13; c = +30;
Δ = b2-4ac
Δ = -132-4·1·30
Δ = 49
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{49}=7$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-7}{2*1}=\frac{6}{2} =3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+7}{2*1}=\frac{20}{2} =10 $
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