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4(3x+2)+2(5x-6)=5x(2x+4)
We move all terms to the left:
4(3x+2)+2(5x-6)-(5x(2x+4))=0
We multiply parentheses
12x+10x-(5x(2x+4))+8-12=0
We calculate terms in parentheses: -(5x(2x+4)), so:We add all the numbers together, and all the variables
5x(2x+4)
We multiply parentheses
10x^2+20x
Back to the equation:
-(10x^2+20x)
22x-(10x^2+20x)-4=0
We get rid of parentheses
-10x^2+22x-20x-4=0
We add all the numbers together, and all the variables
-10x^2+2x-4=0
a = -10; b = 2; c = -4;
Δ = b2-4ac
Δ = 22-4·(-10)·(-4)
Δ = -156
Delta is less than zero, so there is no solution for the equation
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