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(3x-4)(6x-5)=x
We move all terms to the left:
(3x-4)(6x-5)-(x)=0
We add all the numbers together, and all the variables
-1x+(3x-4)(6x-5)=0
We multiply parentheses ..
(+18x^2-15x-24x+20)-1x=0
We get rid of parentheses
18x^2-15x-24x-1x+20=0
We add all the numbers together, and all the variables
18x^2-40x+20=0
a = 18; b = -40; c = +20;
Δ = b2-4ac
Δ = -402-4·18·20
Δ = 160
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{10}}{2*18}=\frac{40-4\sqrt{10}}{36} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{10}}{2*18}=\frac{40+4\sqrt{10}}{36} $
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