5m^2-95=0

Solution for 5m^2-95=0 equation:

5m^2-95=0

a = 5; b = 0; c = -95;
Δ = b2-4ac
Δ = 02-4·5·(-95)
Δ = 1900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1900}=\sqrt{100*19}=\sqrt{100}*\sqrt{19}=10\sqrt{19}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{19}}{2*5}=\frac{0-10\sqrt{19}}{10}
=-\frac{10\sqrt{19}}{10}
=-\sqrt{19}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{19}}{2*5}=\frac{0+10\sqrt{19}}{10}
=\frac{10\sqrt{19}}{10}
=\sqrt{19}
$