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5x(x-4)=3(x+2)
We move all terms to the left:
5x(x-4)-(3(x+2))=0
We multiply parentheses
5x^2-20x-(3(x+2))=0
We calculate terms in parentheses: -(3(x+2)), so:We get rid of parentheses
3(x+2)
We multiply parentheses
3x+6
Back to the equation:
-(3x+6)
5x^2-20x-3x-6=0
We add all the numbers together, and all the variables
5x^2-23x-6=0
a = 5; b = -23; c = -6;
Δ = b2-4ac
Δ = -232-4·5·(-6)
Δ = 649
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{-(-23)-\sqrt{649}}{2*5}=\frac{23-\sqrt{649}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{649}}{2*5}=\frac{23+\sqrt{649}}{10} $
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