2(x-3)(x+3)-(3x+4)=x(2x-5)-18

Solution for 2(x-3)(x+3)-(3x+4)=x(2x-5)-18 equation:

2(x-3)(x+3)-(3x+4)=x(2x-5)-18

We move all terms to the left:

2(x-3)(x+3)-(3x+4)-(x(2x-5)-18)=0

We use the square of the difference formula

x^2-(3x+4)-(x(2x-5)-18)-9=0

We get rid of parentheses

x^2-3x-(x(2x-5)-18)-4-9=0


We calculate terms in parentheses: -(x(2x-5)-18), so:

x(2x-5)-18

We multiply parentheses

2x^2-5x-18

Back to the equation:

-(2x^2-5x-18)


We add all the numbers together, and all the variables

x^2-3x-(2x^2-5x-18)-13=0

We get rid of parentheses

x^2-2x^2-3x+5x+18-13=0

We add all the numbers together, and all the variables

-1x^2+2x+5=0

a = -1; b = 2; c = +5;
Δ = b2-4ac
Δ = 22-4·(-1)·5
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{6}}{2*-1}=\frac{-2-2\sqrt{6}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{6}}{2*-1}=\frac{-2+2\sqrt{6}}{-2}
$