-4.9x^2+19x+300=0

Solution for -4.9x^2+19x+300=0 equation:

-4.9x^2+19x+300=0

a = -4.9; b = 19; c = +300;
Δ = b2-4ac
Δ = 192-4·(-4.9)·300
Δ = 6241
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6241}=79$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-79}{2*-4.9}=\frac{-98}{-9.8}
=+10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+79}{2*-4.9}=\frac{60}{-9.8}
=-6+1/8.1666666666667
$