3x(x-5)-2(x+2)(x-2)=(x+4)(x+1)-16

Solution for 3x(x-5)-2(x+2)(x-2)=(x+4)(x+1)-16 equation:

3x(x-5)-2(x+2)(x-2)=(x+4)(x+1)-16

We move all terms to the left:

3x(x-5)-2(x+2)(x-2)-((x+4)(x+1)-16)=0

We use the square of the difference formula

x^2+3x(x-5)-((x+4)(x+1)-16)+4=0

We multiply parentheses

x^2+3x^2-15x-((x+4)(x+1)-16)+4=0

We multiply parentheses ..

x^2+3x^2-((+x^2+x+4x+4)-16)-15x+4=0


We calculate terms in parentheses: -((+x^2+x+4x+4)-16), so:

(+x^2+x+4x+4)-16

We get rid of parentheses

x^2+x+4x+4-16

We add all the numbers together, and all the variables

x^2+5x-12

Back to the equation:

-(x^2+5x-12)


We add all the numbers together, and all the variables

4x^2-15x-(x^2+5x-12)+4=0

We get rid of parentheses

4x^2-x^2-15x-5x+12+4=0

We add all the numbers together, and all the variables

3x^2-20x+16=0

a = 3; b = -20; c = +16;
Δ = b2-4ac
Δ = -202-4·3·16
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{13}}{2*3}=\frac{20-4\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{13}}{2*3}=\frac{20+4\sqrt{13}}{6}
$