2a-5=4(a+1)9a+3

Solution for 2a-5=4(a+1)9a+3 equation:

2a-5=4(a+1)9a+3

We move all terms to the left:

2a-5-(4(a+1)9a+3)=0


We calculate terms in parentheses: -(4(a+1)9a+3), so:

4(a+1)9a+3

We multiply parentheses

36a^2+36a+3

Back to the equation:

-(36a^2+36a+3)


We get rid of parentheses

-36a^2+2a-36a-3-5=0

We add all the numbers together, and all the variables

-36a^2-34a-8=0

a = -36; b = -34; c = -8;
Δ = b2-4ac
Δ = -342-4·(-36)·(-8)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-2}{2*-36}=\frac{32}{-72}
=-4/9
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+2}{2*-36}=\frac{36}{-72}
=-1/2
$