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4x(x-3)=4x-12
We move all terms to the left:
4x(x-3)-(4x-12)=0
We multiply parentheses
4x^2-12x-(4x-12)=0
We get rid of parentheses
4x^2-12x-4x+12=0
We add all the numbers together, and all the variables
4x^2-16x+12=0
a = 4; b = -16; c = +12;
Δ = b2-4ac
Δ = -162-4·4·12
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8}{2*4}=\frac{8}{8} =1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8}{2*4}=\frac{24}{8} =3 $
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