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10x^2-29x+20=0
a = 10; b = -29; c = +20;
Δ = b2-4ac
Δ = -292-4·10·20
Δ = 41
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{-(-29)-\sqrt{41}}{2*10}=\frac{29-\sqrt{41}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+\sqrt{41}}{2*10}=\frac{29+\sqrt{41}}{20} $
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