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

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



2x(x+3)=2(3x+5)-x
We move all terms to the left:
2x(x+3)-(2(3x+5)-x)=0
We multiply parentheses
2x^2+6x-(2(3x+5)-x)=0
We calculate terms in parentheses: -(2(3x+5)-x), so:
2(3x+5)-x
We add all the numbers together, and all the variables
-1x+2(3x+5)
We multiply parentheses
-1x+6x+10
We add all the numbers together, and all the variables
5x+10
Back to the equation:
-(5x+10)
We get rid of parentheses
2x^2+6x-5x-10=0
We add all the numbers together, and all the variables
2x^2+x-10=0
a = 2; b = 1; c = -10;
Δ = b2-4ac
Δ = 12-4·2·(-10)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*2}=\frac{-10}{4} =-2+1/2 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*2}=\frac{8}{4} =2 $

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