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0=(3-4x)(2x+3)
We move all terms to the left:
0-((3-4x)(2x+3))=0
We add all the numbers together, and all the variables
-((-4x+3)(2x+3))+0=0
We add all the numbers together, and all the variables
-((-4x+3)(2x+3))=0
We multiply parentheses ..
-((-8x^2-12x+6x+9))=0
We calculate terms in parentheses: -((-8x^2-12x+6x+9)), so:We get rid of parentheses
(-8x^2-12x+6x+9)
We get rid of parentheses
-8x^2-12x+6x+9
We add all the numbers together, and all the variables
-8x^2-6x+9
Back to the equation:
-(-8x^2-6x+9)
8x^2+6x-9=0
a = 8; b = 6; c = -9;
Δ = b2-4ac
Δ = 62-4·8·(-9)
Δ = 324
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{324}=18$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-18}{2*8}=\frac{-24}{16} =-1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+18}{2*8}=\frac{12}{16} =3/4 $
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