2q^2-q-48=0

Solution for 2q^2-q-48=0 equation:

2q^2-q-48=0

We add all the numbers together, and all the variables

2q^2-1q-48=0

a = 2; b = -1; c = -48;
Δ = b2-4ac
Δ = -12-4·2·(-48)
Δ = 385
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{385}}{2*2}=\frac{1-\sqrt{385}}{4}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{385}}{2*2}=\frac{1+\sqrt{385}}{4}
$