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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)+10x+6=0
We get rid of parentheses
-2x^2-6x+5x+10x+15+6=0
We add all the numbers together, and all the variables
-2x^2+9x+21=0
a = -2; b = 9; c = +21;
Δ = b2-4ac
Δ = 92-4·(-2)·21
Δ = 249
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{249}}{2*-2}=\frac{-9-\sqrt{249}}{-4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{249}}{2*-2}=\frac{-9+\sqrt{249}}{-4} $
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