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14a^2+2a-20=0
a = 14; b = 2; c = -20;
Δ = b2-4ac
Δ = 22-4·14·(-20)
Δ = 1124
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1124}=\sqrt{4*281}=\sqrt{4}*\sqrt{281}=2\sqrt{281}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{281}}{2*14}=\frac{-2-2\sqrt{281}}{28} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{281}}{2*14}=\frac{-2+2\sqrt{281}}{28} $
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