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2x(x-3)=x(x+5)
We move all terms to the left:
2x(x-3)-(x(x+5))=0
We multiply parentheses
2x^2-6x-(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)
2x^2-x^2-6x-5x=0
We add all the numbers together, and all the variables
x^2-11x=0
a = 1; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·1·0
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*1}=\frac{0}{2} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*1}=\frac{22}{2} =11 $
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