60-q-q^2=10+4q

Solution for 60-q-q^2=10+4q equation:

60-q-q^2=10+4q

We move all terms to the left:

60-q-q^2-(10+4q)=0

We add all the numbers together, and all the variables

-q^2-q-(4q+10)+60=0

We add all the numbers together, and all the variables

-1q^2-1q-(4q+10)+60=0

We get rid of parentheses

-1q^2-1q-4q-10+60=0

We add all the numbers together, and all the variables

-1q^2-5q+50=0

a = -1; b = -5; c = +50;
Δ = b2-4ac
Δ = -52-4·(-1)·50
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-15}{2*-1}=\frac{-10}{-2}
=+5
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+15}{2*-1}=\frac{20}{-2}
=-10
$