(x-8)(3x-4)=1

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Solution for (x-8)(3x-4)=1 equation:



(x-8)(3x-4)=1
We move all terms to the left:
(x-8)(3x-4)-(1)=0
We multiply parentheses ..
(+3x^2-4x-24x+32)-1=0
We get rid of parentheses
3x^2-4x-24x+32-1=0
We add all the numbers together, and all the variables
3x^2-28x+31=0
a = 3; b = -28; c = +31;
Δ = b2-4ac
Δ = -282-4·3·31
Δ = 412
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:
$\sqrt{\Delta}=\sqrt{412}=\sqrt{4*103}=\sqrt{4}*\sqrt{103}=2\sqrt{103}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-2\sqrt{103}}{2*3}=\frac{28-2\sqrt{103}}{6} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+2\sqrt{103}}{2*3}=\frac{28+2\sqrt{103}}{6} $

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