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(x-4)(3x-5)=4x
We move all terms to the left:
(x-4)(3x-5)-(4x)=0
We add all the numbers together, and all the variables
-4x+(x-4)(3x-5)=0
We multiply parentheses ..
(+3x^2-5x-12x+20)-4x=0
We get rid of parentheses
3x^2-5x-12x-4x+20=0
We add all the numbers together, and all the variables
3x^2-21x+20=0
a = 3; b = -21; c = +20;
Δ = b2-4ac
Δ = -212-4·3·20
Δ = 201
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-\sqrt{201}}{2*3}=\frac{21-\sqrt{201}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{201}}{2*3}=\frac{21+\sqrt{201}}{6} $
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