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x(x-3)=5(x-3)=0
We move all terms to the left:
x(x-3)-(5(x-3))=0
We multiply parentheses
x^2-3x-(5(x-3))=0
We calculate terms in parentheses: -(5(x-3)), so:We get rid of parentheses
5(x-3)
We multiply parentheses
5x-15
Back to the equation:
-(5x-15)
x^2-3x-5x+15=0
We add all the numbers together, and all the variables
x^2-8x+15=0
a = 1; b = -8; c = +15;
Δ = b2-4ac
Δ = -82-4·1·15
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2}{2*1}=\frac{6}{2} =3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2}{2*1}=\frac{10}{2} =5 $
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