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4x^2-4x=5x^2-12
We move all terms to the left:
4x^2-4x-(5x^2-12)=0
We get rid of parentheses
4x^2-5x^2-4x+12=0
We add all the numbers together, and all the variables
-1x^2-4x+12=0
a = -1; b = -4; c = +12;
Δ = b2-4ac
Δ = -42-4·(-1)·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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*-1}=\frac{-4}{-2} =+2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*-1}=\frac{12}{-2} =-6 $
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