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(19x^2)-68x+47=0
a = 19; b = -68; c = +47;
Δ = b2-4ac
Δ = -682-4·19·47
Δ = 1052
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1052}=\sqrt{4*263}=\sqrt{4}*\sqrt{263}=2\sqrt{263}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-68)-2\sqrt{263}}{2*19}=\frac{68-2\sqrt{263}}{38} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-68)+2\sqrt{263}}{2*19}=\frac{68+2\sqrt{263}}{38} $
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