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(3x-8)(2x-5)=12x
We move all terms to the left:
(3x-8)(2x-5)-(12x)=0
We add all the numbers together, and all the variables
-12x+(3x-8)(2x-5)=0
We multiply parentheses ..
(+6x^2-15x-16x+40)-12x=0
We get rid of parentheses
6x^2-15x-16x-12x+40=0
We add all the numbers together, and all the variables
6x^2-43x+40=0
a = 6; b = -43; c = +40;
Δ = b2-4ac
Δ = -432-4·6·40
Δ = 889
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{-(-43)-\sqrt{889}}{2*6}=\frac{43-\sqrt{889}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-43)+\sqrt{889}}{2*6}=\frac{43+\sqrt{889}}{12} $
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