4x-4*5x+4=9x^2-16

Solution for 4x-4*5x+4=9x^2-16 equation:

4x-4*5x+4=9x^2-16

We move all terms to the left:

4x-4*5x+4-(9x^2-16)=0

Wy multiply elements

4x-20x-(9x^2-16)+4=0

We get rid of parentheses

-9x^2+4x-20x+16+4=0

We add all the numbers together, and all the variables

-9x^2-16x+20=0

a = -9; b = -16; c = +20;
Δ = b2-4ac
Δ = -162-4·(-9)·20
Δ = 976
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:

$\sqrt{\Delta}=\sqrt{976}=\sqrt{16*61}=\sqrt{16}*\sqrt{61}=4\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{61}}{2*-9}=\frac{16-4\sqrt{61}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{61}}{2*-9}=\frac{16+4\sqrt{61}}{-18}
$