-56m^2+65=0

Solution for -56m^2+65=0 equation:

-56m^2+65=0

a = -56; b = 0; c = +65;
Δ = b2-4ac
Δ = 02-4·(-56)·65
Δ = 14560
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:

$\sqrt{\Delta}=\sqrt{14560}=\sqrt{16*910}=\sqrt{16}*\sqrt{910}=4\sqrt{910}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{910}}{2*-56}=\frac{0-4\sqrt{910}}{-112}
=-\frac{4\sqrt{910}}{-112}
=-\frac{\sqrt{910}}{-28}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{910}}{2*-56}=\frac{0+4\sqrt{910}}{-112}
=\frac{4\sqrt{910}}{-112}
=\frac{\sqrt{910}}{-28}
$