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126=x(2x-4)
We move all terms to the left:
126-(x(2x-4))=0
We calculate terms in parentheses: -(x(2x-4)), so:We get rid of parentheses
x(2x-4)
We multiply parentheses
2x^2-4x
Back to the equation:
-(2x^2-4x)
-2x^2+4x+126=0
a = -2; b = 4; c = +126;
Δ = b2-4ac
Δ = 42-4·(-2)·126
Δ = 1024
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{1024}=32$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-32}{2*-2}=\frac{-36}{-4} =+9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+32}{2*-2}=\frac{28}{-4} =-7 $
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