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(3x-4)(2x-5)=18
We move all terms to the left:
(3x-4)(2x-5)-(18)=0
We multiply parentheses ..
(+6x^2-15x-8x+20)-18=0
We get rid of parentheses
6x^2-15x-8x+20-18=0
We add all the numbers together, and all the variables
6x^2-23x+2=0
a = 6; b = -23; c = +2;
Δ = b2-4ac
Δ = -232-4·6·2
Δ = 481
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{-(-23)-\sqrt{481}}{2*6}=\frac{23-\sqrt{481}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{481}}{2*6}=\frac{23+\sqrt{481}}{12} $
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