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6-6x=3x^2-8x+3
We move all terms to the left:
6-6x-(3x^2-8x+3)=0
We get rid of parentheses
-3x^2-6x+8x-3+6=0
We add all the numbers together, and all the variables
-3x^2+2x+3=0
a = -3; b = 2; c = +3;
Δ = b2-4ac
Δ = 22-4·(-3)·3
Δ = 40
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{10}}{2*-3}=\frac{-2-2\sqrt{10}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{10}}{2*-3}=\frac{-2+2\sqrt{10}}{-6} $
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