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r2+10r+9=0
We add all the numbers together, and all the variables
r^2+10r+9=0
a = 1; b = 10; c = +9;
Δ = b2-4ac
Δ = 102-4·1·9
Δ = 64
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{64}=8$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-8}{2*1}=\frac{-18}{2} =-9 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+8}{2*1}=\frac{-2}{2} =-1 $
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