(x^2-7x+12)=(x+6)

Solution for (x^2-7x+12)=(x+6) equation:

(x^2-7x+12)=(x+6)

We move all terms to the left:

(x^2-7x+12)-((x+6))=0

We get rid of parentheses

x^2-7x-((x+6))+12=0


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

(x+6)

We get rid of parentheses

x+6

Back to the equation:

-(x+6)


We get rid of parentheses

x^2-7x-x-6+12=0

We add all the numbers together, and all the variables

x^2-8x+6=0

a = 1; b = -8; c = +6;
Δ = b2-4ac
Δ = -82-4·1·6
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{10}}{2*1}=\frac{8-2\sqrt{10}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{10}}{2*1}=\frac{8+2\sqrt{10}}{2}
$