q^2+20q=-4q^2+10q+840

Solution for q^2+20q=-4q^2+10q+840 equation:

q^2+20q=-4q^2+10q+840

We move all terms to the left:

q^2+20q-(-4q^2+10q+840)=0

We get rid of parentheses

q^2+4q^2-10q+20q-840=0

We add all the numbers together, and all the variables

5q^2+10q-840=0

a = 5; b = 10; c = -840;
Δ = b2-4ac
Δ = 102-4·5·(-840)
Δ = 16900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16900}=130$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-130}{2*5}=\frac{-140}{10}
=-14
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+130}{2*5}=\frac{120}{10}
=12
$