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3x^2+6x=32+2x
We move all terms to the left:
3x^2+6x-(32+2x)=0
We add all the numbers together, and all the variables
3x^2+6x-(2x+32)=0
We get rid of parentheses
3x^2+6x-2x-32=0
We add all the numbers together, and all the variables
3x^2+4x-32=0
a = 3; b = 4; c = -32;
Δ = b2-4ac
Δ = 42-4·3·(-32)
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20}{2*3}=\frac{-24}{6} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*3}=\frac{16}{6} =2+2/3 $
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