2q^2-7=0

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

2q^2-7=0

a = 2; b = 0; c = -7;
Δ = b2-4ac
Δ = 02-4·2·(-7)
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{14}}{2*2}=\frac{0-2\sqrt{14}}{4}
=-\frac{2\sqrt{14}}{4}
=-\frac{\sqrt{14}}{2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{14}}{2*2}=\frac{0+2\sqrt{14}}{4}
=\frac{2\sqrt{14}}{4}
=\frac{\sqrt{14}}{2}
$