2q^2+8q-3=0

Solution for 2q^2+8q-3=0 equation:

2q^2+8q-3=0

a = 2; b = 8; c = -3;
Δ = b2-4ac
Δ = 82-4·2·(-3)
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{22}}{2*2}=\frac{-8-2\sqrt{22}}{4}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{22}}{2*2}=\frac{-8+2\sqrt{22}}{4}
$