-4.9x^2+40x+25=0

Solution for -4.9x^2+40x+25=0 equation:

-4.9x^2+40x+25=0

a = -4.9; b = 40; c = +25;
Δ = b2-4ac
Δ = 402-4·(-4.9)·25
Δ = 2090
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{-(40)-\sqrt{2090}}{2*-4.9}=\frac{-40-\sqrt{2090}}{-9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+\sqrt{2090}}{2*-4.9}=\frac{-40+\sqrt{2090}}{-9.8}
$