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11x^2-10x-5=0
a = 11; b = -10; c = -5;
Δ = b2-4ac
Δ = -102-4·11·(-5)
Δ = 320
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-8\sqrt{5}}{2*11}=\frac{10-8\sqrt{5}}{22} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+8\sqrt{5}}{2*11}=\frac{10+8\sqrt{5}}{22} $
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