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2x-(3-5x)=4x(4x+3)
We move all terms to the left:
2x-(3-5x)-(4x(4x+3))=0
We add all the numbers together, and all the variables
2x-(-5x+3)-(4x(4x+3))=0
We get rid of parentheses
2x+5x-(4x(4x+3))-3=0
We calculate terms in parentheses: -(4x(4x+3)), so:We add all the numbers together, and all the variables
4x(4x+3)
We multiply parentheses
16x^2+12x
Back to the equation:
-(16x^2+12x)
7x-(16x^2+12x)-3=0
We get rid of parentheses
-16x^2+7x-12x-3=0
We add all the numbers together, and all the variables
-16x^2-5x-3=0
a = -16; b = -5; c = -3;
Δ = b2-4ac
Δ = -52-4·(-16)·(-3)
Δ = -167
Delta is less than zero, so there is no solution for the equation
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