(30+x)(x+100)=1500

Solution for (30+x)(x+100)=1500 equation:

(30+x)(x+100)=1500

We move all terms to the left:

(30+x)(x+100)-(1500)=0

We add all the numbers together, and all the variables

(x+30)(x+100)-1500=0

We multiply parentheses ..

(+x^2+100x+30x+3000)-1500=0

We get rid of parentheses

x^2+100x+30x+3000-1500=0

We add all the numbers together, and all the variables

x^2+130x+1500=0

a = 1; b = 130; c = +1500;
Δ = b2-4ac
Δ = 1302-4·1·1500
Δ = 10900
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:

$\sqrt{\Delta}=\sqrt{10900}=\sqrt{100*109}=\sqrt{100}*\sqrt{109}=10\sqrt{109}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(130)-10\sqrt{109}}{2*1}=\frac{-130-10\sqrt{109}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(130)+10\sqrt{109}}{2*1}=\frac{-130+10\sqrt{109}}{2}
$