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3x^2+14x=420
We move all terms to the left:
3x^2+14x-(420)=0
a = 3; b = 14; c = -420;
Δ = b2-4ac
Δ = 142-4·3·(-420)
Δ = 5236
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5236}=\sqrt{4*1309}=\sqrt{4}*\sqrt{1309}=2\sqrt{1309}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{1309}}{2*3}=\frac{-14-2\sqrt{1309}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{1309}}{2*3}=\frac{-14+2\sqrt{1309}}{6} $
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