x^2+350x+2500=0

Solution for x^2+350x+2500=0 equation:

x^2+350x+2500=0

a = 1; b = 350; c = +2500;
Δ = b2-4ac
Δ = 3502-4·1·2500
Δ = 112500
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:

$\sqrt{\Delta}=\sqrt{112500}=\sqrt{22500*5}=\sqrt{22500}*\sqrt{5}=150\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(350)-150\sqrt{5}}{2*1}=\frac{-350-150\sqrt{5}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(350)+150\sqrt{5}}{2*1}=\frac{-350+150\sqrt{5}}{2}
$