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x^2-29x+168=0
a = 1; b = -29; c = +168;
Δ = b2-4ac
Δ = -292-4·1·168
Δ = 169
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-13}{2*1}=\frac{16}{2} =8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+13}{2*1}=\frac{42}{2} =21 $
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