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26+15r+r^2=0
a = 1; b = 15; c = +26;
Δ = b2-4ac
Δ = 152-4·1·26
Δ = 121
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{121}=11$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-11}{2*1}=\frac{-26}{2} =-13 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+11}{2*1}=\frac{-4}{2} =-2 $
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