(x-7)=(x^2-6x+3)

Solution for (x-7)=(x^2-6x+3) equation:

(x-7)=(x^2-6x+3)

We move all terms to the left:

(x-7)-((x^2-6x+3))=0

We get rid of parentheses

x-((x^2-6x+3))-7=0


We calculate terms in parentheses: -((x^2-6x+3)), so:

(x^2-6x+3)

We get rid of parentheses

x^2-6x+3

Back to the equation:

-(x^2-6x+3)


We get rid of parentheses

-x^2+x+6x-3-7=0

We add all the numbers together, and all the variables

-1x^2+7x-10=0

a = -1; b = 7; c = -10;
Δ = b2-4ac
Δ = 72-4·(-1)·(-10)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-3}{2*-1}=\frac{-10}{-2}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+3}{2*-1}=\frac{-4}{-2}
=+2
$