2q^2+6q-7=0

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

2q^2+6q-7=0

a = 2; b = 6; c = -7;
Δ = b2-4ac
Δ = 62-4·2·(-7)
Δ = 92
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:

$\sqrt{\Delta}=\sqrt{92}=\sqrt{4*23}=\sqrt{4}*\sqrt{23}=2\sqrt{23}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{23}}{2*2}=\frac{-6-2\sqrt{23}}{4}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{23}}{2*2}=\frac{-6+2\sqrt{23}}{4}
$