.1x^2+14x-2400=0

Solution for .1x^2+14x-2400=0 equation:

.1x^2+14x-2400=0

a = .1; b = 14; c = -2400;
Δ = b2-4ac
Δ = 142-4·.1·(-2400)
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-34}{2*.1}=\frac{-48}{0.2}
=-240
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+34}{2*.1}=\frac{20}{0.2}
=100
$