a(23+a)=4(5a+7)

Solution for a(23+a)=4(5a+7) equation:

a(23+a)=4(5a+7)

We move all terms to the left:

a(23+a)-(4(5a+7))=0

We add all the numbers together, and all the variables

a(a+23)-(4(5a+7))=0

We multiply parentheses

a^2+23a-(4(5a+7))=0


We calculate terms in parentheses: -(4(5a+7)), so:

4(5a+7)

We multiply parentheses

20a+28

Back to the equation:

-(20a+28)


We get rid of parentheses

a^2+23a-20a-28=0

We add all the numbers together, and all the variables

a^2+3a-28=0

a = 1; b = 3; c = -28;
Δ = b2-4ac
Δ = 32-4·1·(-28)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-11}{2*1}=\frac{-14}{2}
=-7
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+11}{2*1}=\frac{8}{2}
=4
$