(2x+30)(2x+50)-1500=600

Solution for (2x+30)(2x+50)-1500=600 equation:

(2x+30)(2x+50)-1500=600

We move all terms to the left:

(2x+30)(2x+50)-1500-(600)=0

We add all the numbers together, and all the variables

(2x+30)(2x+50)-2100=0

We multiply parentheses ..

(+4x^2+100x+60x+1500)-2100=0

We get rid of parentheses

4x^2+100x+60x+1500-2100=0

We add all the numbers together, and all the variables

4x^2+160x-600=0

a = 4; b = 160; c = -600;
Δ = b2-4ac
Δ = 1602-4·4·(-600)
Δ = 35200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{35200}=\sqrt{1600*22}=\sqrt{1600}*\sqrt{22}=40\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-40\sqrt{22}}{2*4}=\frac{-160-40\sqrt{22}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+40\sqrt{22}}{2*4}=\frac{-160+40\sqrt{22}}{8}
$