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4x(x-3)+2x(x-2)=9x^2-12x+4-8
We move all terms to the left:
4x(x-3)+2x(x-2)-(9x^2-12x+4-8)=0
We multiply parentheses
4x^2+2x^2-12x-4x-(9x^2-12x+4-8)=0
We get rid of parentheses
4x^2+2x^2-9x^2-12x-4x+12x-4+8=0
We add all the numbers together, and all the variables
-3x^2-4x+4=0
a = -3; b = -4; c = +4;
Δ = b2-4ac
Δ = -42-4·(-3)·4
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*-3}=\frac{-4}{-6} =2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*-3}=\frac{12}{-6} =-2 $
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