-(x+2)(x-5)(x-3)(x+3)=0

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Solution for -(x+2)(x-5)(x-3)(x+3)=0 equation: 



-(x+2)(x-5)(x-3)(x+3)=0

We multiply parentheses ..

-(+x^2-5x+2x-10)(x-3)(x+3)=0

We use the square of the difference formula

x^2+9=0

a = 1; b = 0; c = +9;
Δ = b2-4ac
Δ = 02-4·1·9
Δ = -36
Delta is less than zero, so there is no solution for the equation

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