1500=(2x+30)(2x+10)

Solution for 1500=(2x+30)(2x+10) equation:

1500=(2x+30)(2x+10)

We move all terms to the left:

1500-((2x+30)(2x+10))=0

We multiply parentheses ..

-((+4x^2+20x+60x+300))+1500=0


We calculate terms in parentheses: -((+4x^2+20x+60x+300)), so:

(+4x^2+20x+60x+300)

We get rid of parentheses

4x^2+20x+60x+300

We add all the numbers together, and all the variables

4x^2+80x+300

Back to the equation:

-(4x^2+80x+300)


We get rid of parentheses

-4x^2-80x-300+1500=0

We add all the numbers together, and all the variables

-4x^2-80x+1200=0

a = -4; b = -80; c = +1200;
Δ = b2-4ac
Δ = -802-4·(-4)·1200
Δ = 25600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25600}=160$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-160}{2*-4}=\frac{-80}{-8}
=+10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+160}{2*-4}=\frac{240}{-8}
=-30
$