23=(3a+5)(a-5)

Solution for 23=(3a+5)(a-5) equation:

23=(3a+5)(a-5)

We move all terms to the left:

23-((3a+5)(a-5))=0

We multiply parentheses ..

-((+3a^2-15a+5a-25))+23=0


We calculate terms in parentheses: -((+3a^2-15a+5a-25)), so:

(+3a^2-15a+5a-25)

We get rid of parentheses

3a^2-15a+5a-25

We add all the numbers together, and all the variables

3a^2-10a-25

Back to the equation:

-(3a^2-10a-25)


We get rid of parentheses

-3a^2+10a+25+23=0

We add all the numbers together, and all the variables

-3a^2+10a+48=0

a = -3; b = 10; c = +48;
Δ = b2-4ac
Δ = 102-4·(-3)·48
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-26}{2*-3}=\frac{-36}{-6}
=+6
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+26}{2*-3}=\frac{16}{-6}
=-2+2/3
$