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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

(x+2)(x-5)-(7x-5)=0

We get rid of parentheses

(x+2)(x-5)-7x+5=0

We multiply parentheses ..

(+x^2-5x+2x-10)-7x+5=0

We get rid of parentheses

x^2-5x+2x-7x-10+5=0

We add all the numbers together, and all the variables

x^2-10x-5=0

a = 1; b = -10; c = -5;
Δ = b2-4ac
Δ = -102-4·1·(-5)
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{30}}{2*1}=\frac{10-2\sqrt{30}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{30}}{2*1}=\frac{10+2\sqrt{30}}{2}
$