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3x(x-4)-(3-4)=6x
We move all terms to the left:
3x(x-4)-(3-4)-(6x)=0
We add all the numbers together, and all the variables
3x(x-4)-6x-(-1)=0
We add all the numbers together, and all the variables
-6x+3x(x-4)+1=0
We multiply parentheses
3x^2-6x-12x+1=0
We add all the numbers together, and all the variables
3x^2-18x+1=0
a = 3; b = -18; c = +1;
Δ = b2-4ac
Δ = -182-4·3·1
Δ = 312
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{312}=\sqrt{4*78}=\sqrt{4}*\sqrt{78}=2\sqrt{78}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{78}}{2*3}=\frac{18-2\sqrt{78}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{78}}{2*3}=\frac{18+2\sqrt{78}}{6} $
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