q2+10=7q

Solution for q2+10=7q equation:

q2+10=7q

We move all terms to the left:

q2+10-(7q)=0

We add all the numbers together, and all the variables

q^2-7q+10=0

a = 1; b = -7; c = +10;
Δ = b2-4ac
Δ = -72-4·1·10
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-3}{2*1}=\frac{4}{2}
=2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+3}{2*1}=\frac{10}{2}
=5
$