-(9x+7x)+(3-x)(2x+1)=17

Solution for -(9x+7x)+(3-x)(2x+1)=17 equation:

-(9x+7x)+(3-x)(2x+1)=17

We move all terms to the left:

-(9x+7x)+(3-x)(2x+1)-(17)=0

We add all the numbers together, and all the variables

-(+16x)+(-1x+3)(2x+1)-17=0

We get rid of parentheses

-16x+(-1x+3)(2x+1)-17=0

We multiply parentheses ..

(-2x^2-1x+6x+3)-16x-17=0

We get rid of parentheses

-2x^2-1x+6x-16x+3-17=0

We add all the numbers together, and all the variables

-2x^2-11x-14=0

a = -2; b = -11; c = -14;
Δ = b2-4ac
Δ = -112-4·(-2)·(-14)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-3}{2*-2}=\frac{8}{-4}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+3}{2*-2}=\frac{14}{-4}
=-3+1/2
$