-48x^2+10=0

Solution for -48x^2+10=0 equation:

-48x^2+10=0

a = -48; b = 0; c = +10;
Δ = b2-4ac
Δ = 02-4·(-48)·10
Δ = 1920
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1920}=\sqrt{64*30}=\sqrt{64}*\sqrt{30}=8\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{30}}{2*-48}=\frac{0-8\sqrt{30}}{-96}
=-\frac{8\sqrt{30}}{-96}
=-\frac{\sqrt{30}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{30}}{2*-48}=\frac{0+8\sqrt{30}}{-96}
=\frac{8\sqrt{30}}{-96}
=\frac{\sqrt{30}}{-12}
$