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(x-3)(3x-11)=15
We move all terms to the left:
(x-3)(3x-11)-(15)=0
We multiply parentheses ..
(+3x^2-11x-9x+33)-15=0
We get rid of parentheses
3x^2-11x-9x+33-15=0
We add all the numbers together, and all the variables
3x^2-20x+18=0
a = 3; b = -20; c = +18;
Δ = b2-4ac
Δ = -202-4·3·18
Δ = 184
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{46}}{2*3}=\frac{20-2\sqrt{46}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{46}}{2*3}=\frac{20+2\sqrt{46}}{6} $
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