x^2+51x-1602=0

Solution for x^2+51x-1602=0 equation:

x^2+51x-1602=0

a = 1; b = 51; c = -1602;
Δ = b2-4ac
Δ = 512-4·1·(-1602)
Δ = 9009
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:

$\sqrt{\Delta}=\sqrt{9009}=\sqrt{9*1001}=\sqrt{9}*\sqrt{1001}=3\sqrt{1001}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(51)-3\sqrt{1001}}{2*1}=\frac{-51-3\sqrt{1001}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+3\sqrt{1001}}{2*1}=\frac{-51+3\sqrt{1001}}{2}
$