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

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

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

We move all terms to the left:

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

We get rid of parentheses

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


We calculate terms in parentheses: -((x^2-7x+12)), so:

(x^2-7x+12)

We get rid of parentheses

x^2-7x+12

Back to the equation:

-(x^2-7x+12)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^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}
$