-q^2+35=q^2+5q+2

Solution for -q^2+35=q^2+5q+2 equation:

-q^2+35=q^2+5q+2

We move all terms to the left:

-q^2+35-(q^2+5q+2)=0

We add all the numbers together, and all the variables

-1q^2-(q^2+5q+2)+35=0

We get rid of parentheses

-1q^2-q^2-5q-2+35=0

We add all the numbers together, and all the variables

-2q^2-5q+33=0

a = -2; b = -5; c = +33;
Δ = b2-4ac
Δ = -52-4·(-2)·33
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-17}{2*-2}=\frac{-12}{-4}
=+3
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+17}{2*-2}=\frac{22}{-4}
=-5+1/2
$