8q^2+6q-7=0

Solution for 8q^2+6q-7=0 equation:

8q^2+6q-7=0

a = 8; b = 6; c = -7;
Δ = b2-4ac
Δ = 62-4·8·(-7)
Δ = 260
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:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{65}}{2*8}=\frac{-6-2\sqrt{65}}{16}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{65}}{2*8}=\frac{-6+2\sqrt{65}}{16}
$