4m^2+8m=13

Solution for 4m^2+8m=13 equation:

4m^2+8m=13

We move all terms to the left:

4m^2+8m-(13)=0

a = 4; b = 8; c = -13;
Δ = b2-4ac
Δ = 82-4·4·(-13)
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{17}}{2*4}=\frac{-8-4\sqrt{17}}{8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{17}}{2*4}=\frac{-8+4\sqrt{17}}{8}
$