q^2+20q=-2q^2+10q+15400

Solution for q^2+20q=-2q^2+10q+15400 equation:

q^2+20q=-2q^2+10q+15400

We move all terms to the left:

q^2+20q-(-2q^2+10q+15400)=0

We get rid of parentheses

q^2+2q^2-10q+20q-15400=0

We add all the numbers together, and all the variables

3q^2+10q-15400=0

a = 3; b = 10; c = -15400;
Δ = b2-4ac
Δ = 102-4·3·(-15400)
Δ = 184900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{184900}=430$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-430}{2*3}=\frac{-440}{6}
=-73+1/3
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+430}{2*3}=\frac{420}{6}
=70
$