x2-6x+20=(x-3)2+3x-1

Solution for x2-6x+20=(x-3)2+3x-1 equation:

x2-6x+20=(x-3)2+3x-1

We move all terms to the left:

x2-6x+20-((x-3)2+3x-1)=0

We add all the numbers together, and all the variables

x^2-6x-((x-3)2+3x-1)+20=0


We calculate terms in parentheses: -((x-3)2+3x-1), so:

(x-3)2+3x-1

We add all the numbers together, and all the variables

3x+(x-3)2-1

We multiply parentheses

3x+2x-6-1

We add all the numbers together, and all the variables

5x-7

Back to the equation:

-(5x-7)


We get rid of parentheses

x^2-6x-5x+7+20=0

We add all the numbers together, and all the variables

x^2-11x+27=0

a = 1; b = -11; c = +27;
Δ = b2-4ac
Δ = -112-4·1·27
Δ = 13
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{13}}{2*1}=\frac{11-\sqrt{13}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{13}}{2*1}=\frac{11+\sqrt{13}}{2}
$