10=q2+2q-63=0

Solution for 10=q2+2q-63=0 equation:

10=q2+2q-63=0

We move all terms to the left:

10-(q2+2q-63)=0

We add all the numbers together, and all the variables

-(+q^2+2q-63)+10=0

We get rid of parentheses

-q^2-2q+63+10=0

We add all the numbers together, and all the variables

-1q^2-2q+73=0

a = -1; b = -2; c = +73;
Δ = b2-4ac
Δ = -22-4·(-1)·73
Δ = 296
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:

$\sqrt{\Delta}=\sqrt{296}=\sqrt{4*74}=\sqrt{4}*\sqrt{74}=2\sqrt{74}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{74}}{2*-1}=\frac{2-2\sqrt{74}}{-2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{74}}{2*-1}=\frac{2+2\sqrt{74}}{-2}
$