(x+6)(x-6)=10x+(x-4)2

Solution for (x+6)(x-6)=10x+(x-4)2 equation:

(x+6)(x-6)=10x+(x-4)2

We move all terms to the left:

(x+6)(x-6)-(10x+(x-4)2)=0

We use the square of the difference formula

x^2-(10x+(x-4)2)-36=0


We calculate terms in parentheses: -(10x+(x-4)2), so:

10x+(x-4)2

We multiply parentheses

10x+2x-8

We add all the numbers together, and all the variables

12x-8

Back to the equation:

-(12x-8)


We get rid of parentheses

x^2-12x+8-36=0

We add all the numbers together, and all the variables

x^2-12x-28=0

a = 1; b = -12; c = -28;
Δ = b2-4ac
Δ = -122-4·1·(-28)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-16}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+16}{2*1}=\frac{28}{2}
=14
$