-4.9x^2+100x+25=0

Solution for -4.9x^2+100x+25=0 equation:

-4.9x^2+100x+25=0

a = -4.9; b = 100; c = +25;
Δ = b2-4ac
Δ = 1002-4·(-4.9)·25
Δ = 10490
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{-(100)-\sqrt{10490}}{2*-4.9}=\frac{-100-\sqrt{10490}}{-9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+\sqrt{10490}}{2*-4.9}=\frac{-100+\sqrt{10490}}{-9.8}
$