3x-9^2=(2x+2)(2x+29)

Solution for 3x-9^2=(2x+2)(2x+29) equation:

3x-9^2=(2x+2)(2x+29)

We move all terms to the left:

3x-9^2-((2x+2)(2x+29))=0

We add all the numbers together, and all the variables

3x-((2x+2)(2x+29))-81=0

We multiply parentheses ..

-((+4x^2+58x+4x+58))+3x-81=0


We calculate terms in parentheses: -((+4x^2+58x+4x+58)), so:

(+4x^2+58x+4x+58)

We get rid of parentheses

4x^2+58x+4x+58

We add all the numbers together, and all the variables

4x^2+62x+58

Back to the equation:

-(4x^2+62x+58)


We add all the numbers together, and all the variables

3x-(4x^2+62x+58)-81=0

We get rid of parentheses

-4x^2+3x-62x-58-81=0

We add all the numbers together, and all the variables

-4x^2-59x-139=0

a = -4; b = -59; c = -139;
Δ = b2-4ac
Δ = -592-4·(-4)·(-139)
Δ = 1257
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{-(-59)-\sqrt{1257}}{2*-4}=\frac{59-\sqrt{1257}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-59)+\sqrt{1257}}{2*-4}=\frac{59+\sqrt{1257}}{-8}
$