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3x^2+16x+10=3x
We move all terms to the left:
3x^2+16x+10-(3x)=0
We add all the numbers together, and all the variables
3x^2+13x+10=0
a = 3; b = 13; c = +10;
Δ = b2-4ac
Δ = 132-4·3·10
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-7}{2*3}=\frac{-20}{6} =-3+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+7}{2*3}=\frac{-6}{6} =-1 $
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