(x-5)(x-4)=7(x+9)

Solution for (x-5)(x-4)=7(x+9) equation:

(x-5)(x-4)=7(x+9)

We move all terms to the left:

(x-5)(x-4)-(7(x+9))=0

We multiply parentheses ..

(+x^2-4x-5x+20)-(7(x+9))=0


We calculate terms in parentheses: -(7(x+9)), so:

7(x+9)

We multiply parentheses

7x+63

Back to the equation:

-(7x+63)


We get rid of parentheses

x^2-4x-5x-7x+20-63=0

We add all the numbers together, and all the variables

x^2-16x-43=0

a = 1; b = -16; c = -43;
Δ = b2-4ac
Δ = -162-4·1·(-43)
Δ = 428
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:

$\sqrt{\Delta}=\sqrt{428}=\sqrt{4*107}=\sqrt{4}*\sqrt{107}=2\sqrt{107}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{107}}{2*1}=\frac{16-2\sqrt{107}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{107}}{2*1}=\frac{16+2\sqrt{107}}{2}
$