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3x^2+21x+32=12
We move all terms to the left:
3x^2+21x+32-(12)=0
We add all the numbers together, and all the variables
3x^2+21x+20=0
a = 3; b = 21; c = +20;
Δ = b2-4ac
Δ = 212-4·3·20
Δ = 201
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-\sqrt{201}}{2*3}=\frac{-21-\sqrt{201}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{201}}{2*3}=\frac{-21+\sqrt{201}}{6} $
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