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

Solution for (2x^2-3x-1)-3(2x^2-3x-5)-16=0 equation:

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

We multiply parentheses

-6x^2+(2x^2-3x-1)+9x+15-16=0

We get rid of parentheses

-6x^2+2x^2-3x+9x-1+15-16=0

We add all the numbers together, and all the variables

-4x^2+6x-2=0

a = -4; b = 6; c = -2;
Δ = b2-4ac
Δ = 62-4·(-4)·(-2)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2}{2*-4}=\frac{-8}{-8}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2}{2*-4}=\frac{-4}{-8}
=1/2
$