5(a+3)+4(a+2)a=10

Solution for 5(a+3)+4(a+2)a=10 equation:

5(a+3)+4(a+2)a=10

We move all terms to the left:

5(a+3)+4(a+2)a-(10)=0

We multiply parentheses

4a^2+5a+8a+15-10=0

We add all the numbers together, and all the variables

4a^2+13a+5=0

a = 4; b = 13; c = +5;
Δ = b2-4ac
Δ = 132-4·4·5
Δ = 89
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{89}}{2*4}=\frac{-13-\sqrt{89}}{8}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{89}}{2*4}=\frac{-13+\sqrt{89}}{8}
$