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3x(x-5)+3=3x-11
We move all terms to the left:
3x(x-5)+3-(3x-11)=0
We multiply parentheses
3x^2-15x-(3x-11)+3=0
We get rid of parentheses
3x^2-15x-3x+11+3=0
We add all the numbers together, and all the variables
3x^2-18x+14=0
a = 3; b = -18; c = +14;
Δ = b2-4ac
Δ = -182-4·3·14
Δ = 156
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{39}}{2*3}=\frac{18-2\sqrt{39}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{39}}{2*3}=\frac{18+2\sqrt{39}}{6} $
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