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r^2-20r-29=0
a = 1; b = -20; c = -29;
Δ = b2-4ac
Δ = -202-4·1·(-29)
Δ = 516
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{129}}{2*1}=\frac{20-2\sqrt{129}}{2} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{129}}{2*1}=\frac{20+2\sqrt{129}}{2} $
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