2a(a+4)+6a-10=14

Solution for 2a(a+4)+6a-10=14 equation:

2a(a+4)+6a-10=14

We move all terms to the left:

2a(a+4)+6a-10-(14)=0

We add all the numbers together, and all the variables

6a+2a(a+4)-24=0

We multiply parentheses

2a^2+6a+8a-24=0

We add all the numbers together, and all the variables

2a^2+14a-24=0

a = 2; b = 14; c = -24;
Δ = b2-4ac
Δ = 142-4·2·(-24)
Δ = 388
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:

$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{97}}{2*2}=\frac{-14-2\sqrt{97}}{4}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{97}}{2*2}=\frac{-14+2\sqrt{97}}{4}
$