0=(4x^2+20x+100)-100+9

Solution for 0=(4x^2+20x+100)-100+9 equation:

0=(4x^2+20x+100)-100+9

We move all terms to the left:

0-((4x^2+20x+100)-100+9)=0

We add all the numbers together, and all the variables

-((4x^2+20x+100)-100+9)=0


We calculate terms in parentheses: -((4x^2+20x+100)-100+9), so:

(4x^2+20x+100)-100+9

We add all the numbers together, and all the variables

(4x^2+20x+100)-91

We get rid of parentheses

4x^2+20x+100-91

We add all the numbers together, and all the variables

4x^2+20x+9

Back to the equation:

-(4x^2+20x+9)


We get rid of parentheses

-4x^2-20x-9=0

a = -4; b = -20; c = -9;
Δ = b2-4ac
Δ = -202-4·(-4)·(-9)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-16}{2*-4}=\frac{4}{-8}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+16}{2*-4}=\frac{36}{-8}
=-4+1/2
$