a^2-11a+28=10

Solution for a^2-11a+28=10 equation:

a^2-11a+28=10

We move all terms to the left:

a^2-11a+28-(10)=0

We add all the numbers together, and all the variables

a^2-11a+18=0

a = 1; b = -11; c = +18;
Δ = b2-4ac
Δ = -112-4·1·18
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-7}{2*1}=\frac{4}{2}
=2
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+7}{2*1}=\frac{18}{2}
=9
$