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c^2-4c-20=0
a = 1; b = -4; c = -20;
Δ = b2-4ac
Δ = -42-4·1·(-20)
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{6}}{2*1}=\frac{4-4\sqrt{6}}{2} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{6}}{2*1}=\frac{4+4\sqrt{6}}{2} $
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