(6x-5)=(x^2-3x+2)

Solution for (6x-5)=(x^2-3x+2) equation:

(6x-5)=(x^2-3x+2)

We move all terms to the left:

(6x-5)-((x^2-3x+2))=0

We get rid of parentheses

6x-((x^2-3x+2))-5=0


We calculate terms in parentheses: -((x^2-3x+2)), so:

(x^2-3x+2)

We get rid of parentheses

x^2-3x+2

Back to the equation:

-(x^2-3x+2)


We get rid of parentheses

-x^2+6x+3x-2-5=0

We add all the numbers together, and all the variables

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

a = -1; b = 9; c = -7;
Δ = b2-4ac
Δ = 92-4·(-1)·(-7)
Δ = 53
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{-(9)-\sqrt{53}}{2*-1}=\frac{-9-\sqrt{53}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{53}}{2*-1}=\frac{-9+\sqrt{53}}{-2}
$