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4x^2-12x=16
We move all terms to the left:
4x^2-12x-(16)=0
a = 4; b = -12; c = -16;
Δ = b2-4ac
Δ = -122-4·4·(-16)
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-20}{2*4}=\frac{-8}{8} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+20}{2*4}=\frac{32}{8} =4 $
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