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5x^2+12=3x^2-20
We move all terms to the left:
5x^2+12-(3x^2-20)=0
We get rid of parentheses
5x^2-3x^2+20+12=0
We add all the numbers together, and all the variables
2x^2+32=0
a = 2; b = 0; c = +32;
Δ = b2-4ac
Δ = 02-4·2·32
Δ = -256
Delta is less than zero, so there is no solution for the equation
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