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39x+126=-3x^2
We move all terms to the left:
39x+126-(-3x^2)=0
We get rid of parentheses
3x^2+39x+126=0
a = 3; b = 39; c = +126;
Δ = b2-4ac
Δ = 392-4·3·126
Δ = 9
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-3}{2*3}=\frac{-42}{6} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+3}{2*3}=\frac{-36}{6} =-6 $
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