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11x^2-6x-140=0
a = 11; b = -6; c = -140;
Δ = b2-4ac
Δ = -62-4·11·(-140)
Δ = 6196
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{6196}=\sqrt{4*1549}=\sqrt{4}*\sqrt{1549}=2\sqrt{1549}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{1549}}{2*11}=\frac{6-2\sqrt{1549}}{22} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{1549}}{2*11}=\frac{6+2\sqrt{1549}}{22} $
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