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x^2+129x+190=0
a = 1; b = 129; c = +190;
Δ = b2-4ac
Δ = 1292-4·1·190
Δ = 15881
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(129)-\sqrt{15881}}{2*1}=\frac{-129-\sqrt{15881}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(129)+\sqrt{15881}}{2*1}=\frac{-129+\sqrt{15881}}{2} $
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