x^2+20x-300=3200

Solution for x^2+20x-300=3200 equation:

x^2+20x-300=3200

We move all terms to the left:

x^2+20x-300-(3200)=0

We add all the numbers together, and all the variables

x^2+20x-3500=0

a = 1; b = 20; c = -3500;
Δ = b2-4ac
Δ = 202-4·1·(-3500)
Δ = 14400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{14400}=120$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-120}{2*1}=\frac{-140}{2}
=-70
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+120}{2*1}=\frac{100}{2}
=50
$