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X^2+20X+65=0
a = 1; b = 20; c = +65;
Δ = b2-4ac
Δ = 202-4·1·65
Δ = 140
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{35}}{2*1}=\frac{-20-2\sqrt{35}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{35}}{2*1}=\frac{-20+2\sqrt{35}}{2} $
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