(x^2-32x+252)=-(2x-34)-x+20

Solution for (x^2-32x+252)=-(2x-34)-x+20 equation:

(x^2-32x+252)=-(2x-34)-x+20

We move all terms to the left:

(x^2-32x+252)-(-(2x-34)-x+20)=0

We get rid of parentheses

x^2-32x-(-(2x-34)-x+20)+252=0


We calculate terms in parentheses: -(-(2x-34)-x+20), so:

-(2x-34)-x+20

We add all the numbers together, and all the variables

-1x-(2x-34)+20

We get rid of parentheses

-1x-2x+34+20

We add all the numbers together, and all the variables

-3x+54

Back to the equation:

-(-3x+54)


We get rid of parentheses

x^2-32x+3x-54+252=0

We add all the numbers together, and all the variables

x^2-29x+198=0

a = 1; b = -29; c = +198;
Δ = b2-4ac
Δ = -292-4·1·198
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-7}{2*1}=\frac{22}{2}
=11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+7}{2*1}=\frac{36}{2}
=18
$