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14x^2-10x-20=0
a = 14; b = -10; c = -20;
Δ = b2-4ac
Δ = -102-4·14·(-20)
Δ = 1220
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1220}=\sqrt{4*305}=\sqrt{4}*\sqrt{305}=2\sqrt{305}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{305}}{2*14}=\frac{10-2\sqrt{305}}{28} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{305}}{2*14}=\frac{10+2\sqrt{305}}{28} $
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