(x-5)(x-2)=9x+13

Solution for (x-5)(x-2)=9x+13 equation:

(x-5)(x-2)=9x+13

We move all terms to the left:

(x-5)(x-2)-(9x+13)=0

We get rid of parentheses

(x-5)(x-2)-9x-13=0

We multiply parentheses ..

(+x^2-2x-5x+10)-9x-13=0

We get rid of parentheses

x^2-2x-5x-9x+10-13=0

We add all the numbers together, and all the variables

x^2-16x-3=0

a = 1; b = -16; c = -3;
Δ = b2-4ac
Δ = -162-4·1·(-3)
Δ = 268
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:

$\sqrt{\Delta}=\sqrt{268}=\sqrt{4*67}=\sqrt{4}*\sqrt{67}=2\sqrt{67}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{67}}{2*1}=\frac{16-2\sqrt{67}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{67}}{2*1}=\frac{16+2\sqrt{67}}{2}
$