w^2+5w-1050=0

Solution for w^2+5w-1050=0 equation:

w^2+5w-1050=0

a = 1; b = 5; c = -1050;
Δ = b2-4ac
Δ = 52-4·1·(-1050)
Δ = 4225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4225}=65$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-65}{2*1}=\frac{-70}{2}
=-35
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+65}{2*1}=\frac{60}{2}
=30
$