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5(x-3)=2x(x-3)
We move all terms to the left:
5(x-3)-(2x(x-3))=0
We multiply parentheses
5x-(2x(x-3))-15=0
We calculate terms in parentheses: -(2x(x-3)), so:We get rid of parentheses
2x(x-3)
We multiply parentheses
2x^2-6x
Back to the equation:
-(2x^2-6x)
-2x^2+5x+6x-15=0
We add all the numbers together, and all the variables
-2x^2+11x-15=0
a = -2; b = 11; c = -15;
Δ = b2-4ac
Δ = 112-4·(-2)·(-15)
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-1}{2*-2}=\frac{-12}{-4} =+3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+1}{2*-2}=\frac{-10}{-4} =2+1/2 $
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