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

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

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

We move all terms to the left:

3x-8-(x(x-4)+1)=0


We calculate terms in parentheses: -(x(x-4)+1), so:

x(x-4)+1

We multiply parentheses

x^2-4x+1

Back to the equation:

-(x^2-4x+1)


We get rid of parentheses

-x^2+3x+4x-1-8=0

We add all the numbers together, and all the variables

-1x^2+7x-9=0

a = -1; b = 7; c = -9;
Δ = b2-4ac
Δ = 72-4·(-1)·(-9)
Δ = 13
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{-(7)-\sqrt{13}}{2*-1}=\frac{-7-\sqrt{13}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{13}}{2*-1}=\frac{-7+\sqrt{13}}{-2}
$