(q-6)(7q+1)=0

Solution for (q-6)(7q+1)=0 equation:

(q-6)(7q+1)=0

We multiply parentheses ..

(+7q^2+q-42q-6)=0

We get rid of parentheses

7q^2+q-42q-6=0

We add all the numbers together, and all the variables

7q^2-41q-6=0

a = 7; b = -41; c = -6;
Δ = b2-4ac
Δ = -412-4·7·(-6)
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-43}{2*7}=\frac{-2}{14}
=-1/7
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+43}{2*7}=\frac{84}{14}
=6
$