(a*2)+(a+4)+a(a-)=14

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

(a*2)+(a+4)+a(a-)=14

We move all terms to the left:

(a*2)+(a+4)+a(a-)-(14)=0

We add all the numbers together, and all the variables

(+a*2)+(a+4)+a(+a)-14=0

We multiply parentheses

a^2+(+a*2)+(a+4)-14=0

We get rid of parentheses

a^2+a*2+a+4-14=0

We add all the numbers together, and all the variables

a^2+a+a*2-10=0

Wy multiply elements

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

We add all the numbers together, and all the variables

a^2+3a-10=0

a = 1; b = 3; c = -10;
Δ = b2-4ac
Δ = 32-4·1·(-10)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-7}{2*1}=\frac{-10}{2}
=-5
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*1}=\frac{4}{2}
=2
$