x^2+9x-2142=0

Solution for x^2+9x-2142=0 equation:

x^2+9x-2142=0

a = 1; b = 9; c = -2142;
Δ = b2-4ac
Δ = 92-4·1·(-2142)
Δ = 8649
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8649}=93$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-93}{2*1}=\frac{-102}{2}
=-51
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+93}{2*1}=\frac{84}{2}
=42
$