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

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

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

We move all terms to the left:

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

We get rid of parentheses

q^2+q^2+5q-35+2=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{-22}{4}
=-5+1/2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+17}{2*2}=\frac{12}{4}
=3
$