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r^2+10r-39=0
a = 1; b = 10; c = -39;
Δ = b2-4ac
Δ = 102-4·1·(-39)
Δ = 256
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{256}=16$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-16}{2*1}=\frac{-26}{2} =-13 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+16}{2*1}=\frac{6}{2} =3 $
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