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9x^2+3=-12x
We move all terms to the left:
9x^2+3-(-12x)=0
We get rid of parentheses
9x^2+12x+3=0
a = 9; b = 12; c = +3;
Δ = b2-4ac
Δ = 122-4·9·3
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6}{2*9}=\frac{-18}{18} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6}{2*9}=\frac{-6}{18} =-1/3 $
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