(9x-16)(7x-5)=(3x+11)

Solution for (9x-16)(7x-5)=(3x+11) equation:

(9x-16)(7x-5)=(3x+11)

We move all terms to the left:

(9x-16)(7x-5)-((3x+11))=0

We multiply parentheses ..

(+63x^2-45x-112x+80)-((3x+11))=0


We calculate terms in parentheses: -((3x+11)), so:

(3x+11)

We get rid of parentheses

3x+11

Back to the equation:

-(3x+11)


We get rid of parentheses

63x^2-45x-112x-3x+80-11=0

We add all the numbers together, and all the variables

63x^2-160x+69=0

a = 63; b = -160; c = +69;
Δ = b2-4ac
Δ = -1602-4·63·69
Δ = 8212
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:

$\sqrt{\Delta}=\sqrt{8212}=\sqrt{4*2053}=\sqrt{4}*\sqrt{2053}=2\sqrt{2053}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-2\sqrt{2053}}{2*63}=\frac{160-2\sqrt{2053}}{126}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+2\sqrt{2053}}{2*63}=\frac{160+2\sqrt{2053}}{126}
$