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(2x^2+7x+10)=(x2+2x-9)
We move all terms to the left:
(2x^2+7x+10)-((x2+2x-9))=0
We add all the numbers together, and all the variables
-((+x^2+2x-9))+(2x^2+7x+10)=0
We get rid of parentheses
-((+x^2+2x-9))+2x^2+7x+10=0
We calculate terms in parentheses: -((+x^2+2x-9)), so:We add all the numbers together, and all the variables
(+x^2+2x-9)
We get rid of parentheses
x^2+2x-9
Back to the equation:
-(x^2+2x-9)
2x^2+7x-(x^2+2x-9)+10=0
We get rid of parentheses
2x^2-x^2+7x-2x+9+10=0
We add all the numbers together, and all the variables
x^2+5x+19=0
a = 1; b = 5; c = +19;
Δ = b2-4ac
Δ = 52-4·1·19
Δ = -51
Delta is less than zero, so there is no solution for the equation
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