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11x^2+10x-20=0
a = 11; b = 10; c = -20;
Δ = b2-4ac
Δ = 102-4·11·(-20)
Δ = 980
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{980}=\sqrt{196*5}=\sqrt{196}*\sqrt{5}=14\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-14\sqrt{5}}{2*11}=\frac{-10-14\sqrt{5}}{22} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+14\sqrt{5}}{2*11}=\frac{-10+14\sqrt{5}}{22} $
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