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x^2-35x-50=0
a = 1; b = -35; c = -50;
Δ = b2-4ac
Δ = -352-4·1·(-50)
Δ = 1425
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1425}=\sqrt{25*57}=\sqrt{25}*\sqrt{57}=5\sqrt{57}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-5\sqrt{57}}{2*1}=\frac{35-5\sqrt{57}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+5\sqrt{57}}{2*1}=\frac{35+5\sqrt{57}}{2} $
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