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29x^2+2x-20=0
a = 29; b = 2; c = -20;
Δ = b2-4ac
Δ = 22-4·29·(-20)
Δ = 2324
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2324}=\sqrt{4*581}=\sqrt{4}*\sqrt{581}=2\sqrt{581}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{581}}{2*29}=\frac{-2-2\sqrt{581}}{58} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{581}}{2*29}=\frac{-2+2\sqrt{581}}{58} $
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