0=(a^2-11a+28)

Solution for 0=(a^2-11a+28) equation:

0=(a^2-11a+28)

We move all terms to the left:

0-((a^2-11a+28))=0

We add all the numbers together, and all the variables

-((a^2-11a+28))=0


We calculate terms in parentheses: -((a^2-11a+28)), so:

(a^2-11a+28)

We get rid of parentheses

a^2-11a+28

Back to the equation:

-(a^2-11a+28)


We get rid of parentheses

-a^2+11a-28=0

We add all the numbers together, and all the variables

-1a^2+11a-28=0

a = -1; b = 11; c = -28;
Δ = b2-4ac
Δ = 112-4·(-1)·(-28)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-3}{2*-1}=\frac{-14}{-2}
=+7
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+3}{2*-1}=\frac{-8}{-2}
=+4
$