x^2-10x+27=16

Solution for x^2-10x+27=16 equation:

x^2-10x+27=16

We move all terms to the left:

x^2-10x+27-(16)=0

We add all the numbers together, and all the variables

x^2-10x+11=0

a = 1; b = -10; c = +11;
Δ = b2-4ac
Δ = -102-4·1·11
Δ = 56
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{14}}{2*1}=\frac{10-2\sqrt{14}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{14}}{2*1}=\frac{10+2\sqrt{14}}{2}
$