8m2+20m=2

Solution for 8m2+20m=2 equation:

8m^2+20m=2

We move all terms to the left:

8m^2+20m-(2)=0

a = 8; b = 20; c = -2;
Δ = b2-4ac
Δ = 202-4·8·(-2)
Δ = 464
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:

$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{29}}{2*8}=\frac{-20-4\sqrt{29}}{16}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{29}}{2*8}=\frac{-20+4\sqrt{29}}{16}
$