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3x(6x+4)=(2x-1)(3x+2)
We move all terms to the left:
3x(6x+4)-((2x-1)(3x+2))=0
We multiply parentheses
18x^2+12x-((2x-1)(3x+2))=0
We multiply parentheses ..
18x^2-((+6x^2+4x-3x-2))+12x=0
We calculate terms in parentheses: -((+6x^2+4x-3x-2)), so:We add all the numbers together, and all the variables
(+6x^2+4x-3x-2)
We get rid of parentheses
6x^2+4x-3x-2
We add all the numbers together, and all the variables
6x^2+x-2
Back to the equation:
-(6x^2+x-2)
18x^2+12x-(6x^2+x-2)=0
We get rid of parentheses
18x^2-6x^2+12x-x+2=0
We add all the numbers together, and all the variables
12x^2+11x+2=0
a = 12; b = 11; c = +2;
Δ = b2-4ac
Δ = 112-4·12·2
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-5}{2*12}=\frac{-16}{24} =-2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+5}{2*12}=\frac{-6}{24} =-1/4 $
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