-5x2+242=0

Solution for -5x2+242=0 equation:

-5x^2+242=0

a = -5; b = 0; c = +242;
Δ = b2-4ac
Δ = 02-4·(-5)·242
Δ = 4840
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:

$\sqrt{\Delta}=\sqrt{4840}=\sqrt{484*10}=\sqrt{484}*\sqrt{10}=22\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-22\sqrt{10}}{2*-5}=\frac{0-22\sqrt{10}}{-10}
=-\frac{22\sqrt{10}}{-10}
=-\frac{11\sqrt{10}}{-5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+22\sqrt{10}}{2*-5}=\frac{0+22\sqrt{10}}{-10}
=\frac{22\sqrt{10}}{-10}
=\frac{11\sqrt{10}}{-5}
$