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(x+2)(x-3)-(2x-5)(x+3)=x(x-5)
We move all terms to the left:
(x+2)(x-3)-(2x-5)(x+3)-(x(x-5))=0
We multiply parentheses ..
(+x^2-3x+2x-6)-(2x-5)(x+3)-(x(x-5))=0
We calculate terms in parentheses: -(x(x-5)), so:We get rid of parentheses
x(x-5)
We multiply parentheses
x^2-5x
Back to the equation:
-(x^2-5x)
x^2-x^2-3x+2x-(2x-5)(x+3)+5x-6=0
We multiply parentheses ..
x^2-x^2-(+2x^2+6x-5x-15)-3x+2x+5x-6=0
We add all the numbers together, and all the variables
-(+2x^2+6x-5x-15)+4x-6=0
We get rid of parentheses
-2x^2-6x+5x+4x+15-6=0
We add all the numbers together, and all the variables
-2x^2+3x+9=0
a = -2; b = 3; c = +9;
Δ = b2-4ac
Δ = 32-4·(-2)·9
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*-2}=\frac{-12}{-4} =+3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*-2}=\frac{6}{-4} =-1+1/2 $
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