-4.9x^2+10x+1200=0

Solution for -4.9x^2+10x+1200=0 equation:

-4.9x^2+10x+1200=0

a = -4.9; b = 10; c = +1200;
Δ = b2-4ac
Δ = 102-4·(-4.9)·1200
Δ = 23620
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:

$\sqrt{\Delta}=\sqrt{23620}=\sqrt{4*5905}=\sqrt{4}*\sqrt{5905}=2\sqrt{5905}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{5905}}{2*-4.9}=\frac{-10-2\sqrt{5905}}{-9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{5905}}{2*-4.9}=\frac{-10+2\sqrt{5905}}{-9.8}
$