20=3(x-5)(x-5)+8

Solution for 20=3(x-5)(x-5)+8 equation:

20=3(x-5)(x-5)+8

We move all terms to the left:

20-(3(x-5)(x-5)+8)=0

We multiply parentheses ..

-(3(+x^2-5x-5x+25)+8)+20=0


We calculate terms in parentheses: -(3(+x^2-5x-5x+25)+8), so:

3(+x^2-5x-5x+25)+8

We multiply parentheses

3x^2-15x-15x+75+8

We add all the numbers together, and all the variables

3x^2-30x+83

Back to the equation:

-(3x^2-30x+83)


We get rid of parentheses

-3x^2+30x-83+20=0

We add all the numbers together, and all the variables

-3x^2+30x-63=0

a = -3; b = 30; c = -63;
Δ = b2-4ac
Δ = 302-4·(-3)·(-63)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-12}{2*-3}=\frac{-42}{-6}
=+7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+12}{2*-3}=\frac{-18}{-6}
=+3
$