q^2+10-120=0

Solution for q^2+10-120=0 equation:

q^2+10-120=0

We add all the numbers together, and all the variables

q^2-110=0

a = 1; b = 0; c = -110;
Δ = b2-4ac
Δ = 02-4·1·(-110)
Δ = 440
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:

$\sqrt{\Delta}=\sqrt{440}=\sqrt{4*110}=\sqrt{4}*\sqrt{110}=2\sqrt{110}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{110}}{2*1}=\frac{0-2\sqrt{110}}{2}
=-\frac{2\sqrt{110}}{2}
=-\sqrt{110}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{110}}{2*1}=\frac{0+2\sqrt{110}}{2}
=\frac{2\sqrt{110}}{2}
=\sqrt{110}
$