-4.9x^2+13x+100=0

Solution for -4.9x^2+13x+100=0 equation:

-4.9x^2+13x+100=0

a = -4.9; b = 13; c = +100;
Δ = b2-4ac
Δ = 132-4·(-4.9)·100
Δ = 2129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{2129}}{2*-4.9}=\frac{-13-\sqrt{2129}}{-9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{2129}}{2*-4.9}=\frac{-13+\sqrt{2129}}{-9.8}
$