(10x+8)(5x-4)=12x+16

Solution for (10x+8)(5x-4)=12x+16 equation:

(10x+8)(5x-4)=12x+16

We move all terms to the left:

(10x+8)(5x-4)-(12x+16)=0

We get rid of parentheses

(10x+8)(5x-4)-12x-16=0

We multiply parentheses ..

(+50x^2-40x+40x-32)-12x-16=0

We get rid of parentheses

50x^2-40x+40x-12x-32-16=0

We add all the numbers together, and all the variables

50x^2-12x-48=0

a = 50; b = -12; c = -48;
Δ = b2-4ac
Δ = -122-4·50·(-48)
Δ = 9744
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:

$\sqrt{\Delta}=\sqrt{9744}=\sqrt{16*609}=\sqrt{16}*\sqrt{609}=4\sqrt{609}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{609}}{2*50}=\frac{12-4\sqrt{609}}{100}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{609}}{2*50}=\frac{12+4\sqrt{609}}{100}
$