8=4+(a-5)(a-5)

Solution for 8=4+(a-5)(a-5) equation:

8=4+(a-5)(a-5)

We move all terms to the left:

8-(4+(a-5)(a-5))=0

We multiply parentheses ..

-(4+(+a^2-5a-5a+25))+8=0


We calculate terms in parentheses: -(4+(+a^2-5a-5a+25)), so:

4+(+a^2-5a-5a+25)

determiningTheFunctionDomain
(+a^2-5a-5a+25)+4

We get rid of parentheses

a^2-5a-5a+25+4

We add all the numbers together, and all the variables

a^2-10a+29

Back to the equation:

-(a^2-10a+29)


We get rid of parentheses

-a^2+10a-29+8=0

We add all the numbers together, and all the variables

-1a^2+10a-21=0

a = -1; b = 10; c = -21;
Δ = b2-4ac
Δ = 102-4·(-1)·(-21)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4}{2*-1}=\frac{-14}{-2}
=+7
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4}{2*-1}=\frac{-6}{-2}
=+3
$