2a^2+20a+10=2

Solution for 2a^2+20a+10=2 equation:

2a^2+20a+10=2

We move all terms to the left:

2a^2+20a+10-(2)=0

We add all the numbers together, and all the variables

2a^2+20a+8=0

a = 2; b = 20; c = +8;
Δ = b2-4ac
Δ = 202-4·2·8
Δ = 336
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{21}}{2*2}=\frac{-20-4\sqrt{21}}{4}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{21}}{2*2}=\frac{-20+4\sqrt{21}}{4}
$