2q^2+9=205

Solution for 2q^2+9=205 equation:

2q^2+9=205

We move all terms to the left:

2q^2+9-(205)=0

We add all the numbers together, and all the variables

2q^2-196=0

a = 2; b = 0; c = -196;
Δ = b2-4ac
Δ = 02-4·2·(-196)
Δ = 1568
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:

$\sqrt{\Delta}=\sqrt{1568}=\sqrt{784*2}=\sqrt{784}*\sqrt{2}=28\sqrt{2}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-28\sqrt{2}}{2*2}=\frac{0-28\sqrt{2}}{4}
=-\frac{28\sqrt{2}}{4}
=-7\sqrt{2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+28\sqrt{2}}{2*2}=\frac{0+28\sqrt{2}}{4}
=\frac{28\sqrt{2}}{4}
=7\sqrt{2}
$