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2x(3x-7)+4x(3x+2)=6x(5x+9)
We move all terms to the left:
2x(3x-7)+4x(3x+2)-(6x(5x+9))=0
We multiply parentheses
6x^2+12x^2-14x+8x-(6x(5x+9))=0
We calculate terms in parentheses: -(6x(5x+9)), so:We add all the numbers together, and all the variables
6x(5x+9)
We multiply parentheses
30x^2+54x
Back to the equation:
-(30x^2+54x)
18x^2-6x-(30x^2+54x)=0
We get rid of parentheses
18x^2-30x^2-6x-54x=0
We add all the numbers together, and all the variables
-12x^2-60x=0
a = -12; b = -60; c = 0;
Δ = b2-4ac
Δ = -602-4·(-12)·0
Δ = 3600
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{3600}=60$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-60}{2*-12}=\frac{0}{-24} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+60}{2*-12}=\frac{120}{-24} =-5 $
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