2a+4a(a-5)=10

Solution for 2a+4a(a-5)=10 equation:

2a+4a(a-5)=10

We move all terms to the left:

2a+4a(a-5)-(10)=0

We multiply parentheses

4a^2+2a-20a-10=0

We add all the numbers together, and all the variables

4a^2-18a-10=0

a = 4; b = -18; c = -10;
Δ = b2-4ac
Δ = -182-4·4·(-10)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-22}{2*4}=\frac{-4}{8}
=-1/2
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+22}{2*4}=\frac{40}{8}
=5
$