(16-2x)(21-2x)=64

Solution for (16-2x)(21-2x)=64 equation:

(16-2x)(21-2x)=64

We move all terms to the left:

(16-2x)(21-2x)-(64)=0

We add all the numbers together, and all the variables

(-2x+16)(-2x+21)-64=0

We multiply parentheses ..

(+4x^2-42x-32x+336)-64=0

We get rid of parentheses

4x^2-42x-32x+336-64=0

We add all the numbers together, and all the variables

4x^2-74x+272=0

a = 4; b = -74; c = +272;
Δ = b2-4ac
Δ = -742-4·4·272
Δ = 1124
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:

$\sqrt{\Delta}=\sqrt{1124}=\sqrt{4*281}=\sqrt{4}*\sqrt{281}=2\sqrt{281}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-74)-2\sqrt{281}}{2*4}=\frac{74-2\sqrt{281}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-74)+2\sqrt{281}}{2*4}=\frac{74+2\sqrt{281}}{8}
$