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5^2+2x^2=3x-5^2
We move all terms to the left:
5^2+2x^2-(3x-5^2)=0
We add all the numbers together, and all the variables
2x^2-(3x-5^2)+25=0
We get rid of parentheses
2x^2-3x+25+5^2=0
We add all the numbers together, and all the variables
2x^2-3x+50=0
a = 2; b = -3; c = +50;
Δ = b2-4ac
Δ = -32-4·2·50
Δ = -391
Delta is less than zero, so there is no solution for the equation
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