2m^2+4m-13=0

Solution for 2m^2+4m-13=0 equation:

2m^2+4m-13=0

a = 2; b = 4; c = -13;
Δ = b2-4ac
Δ = 42-4·2·(-13)
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{30}}{2*2}=\frac{-4-2\sqrt{30}}{4}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{30}}{2*2}=\frac{-4+2\sqrt{30}}{4}
$