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4x(3x-1)=16
We move all terms to the left:
4x(3x-1)-(16)=0
We multiply parentheses
12x^2-4x-16=0
a = 12; b = -4; c = -16;
Δ = b2-4ac
Δ = -42-4·12·(-16)
Δ = 784
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{784}=28$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-28}{2*12}=\frac{-24}{24} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+28}{2*12}=\frac{32}{24} =1+1/3 $
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