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3x(2x+1)=3x+2
We move all terms to the left:
3x(2x+1)-(3x+2)=0
We multiply parentheses
6x^2+3x-(3x+2)=0
We get rid of parentheses
6x^2+3x-3x-2=0
We add all the numbers together, and all the variables
6x^2-2=0
a = 6; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·6·(-2)
Δ = 48
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*6}=\frac{0-4\sqrt{3}}{12} =-\frac{4\sqrt{3}}{12} =-\frac{\sqrt{3}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*6}=\frac{0+4\sqrt{3}}{12} =\frac{4\sqrt{3}}{12} =\frac{\sqrt{3}}{3} $
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