q^2+20Q=4Q^2+10q+13000

Solution for q^2+20Q=4Q^2+10q+13000 equation:

q^2+20=4^2+10q+13000

We move all terms to the left:

q^2+20-(4^2+10q+13000)=0

We get rid of parentheses

q^2-10q-13000+20-4^2=0

We add all the numbers together, and all the variables

q^2-10q-12996=0

a = 1; b = -10; c = -12996;
Δ = b2-4ac
Δ = -102-4·1·(-12996)
Δ = 52084
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{52084}=\sqrt{4*13021}=\sqrt{4}*\sqrt{13021}=2\sqrt{13021}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{13021}}{2*1}=\frac{10-2\sqrt{13021}}{2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{13021}}{2*1}=\frac{10+2\sqrt{13021}}{2}
$