-4.9x^2+20x-16=0

Solution for -4.9x^2+20x-16=0 equation:

-4.9x^2+20x-16=0

a = -4.9; b = 20; c = -16;
Δ = b2-4ac
Δ = 202-4·(-4.9)·(-16)
Δ = 86.4
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{-(20)-\sqrt{86.4}}{2*-4.9}=\frac{-20-\sqrt{86.4}}{-9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+\sqrt{86.4}}{2*-4.9}=\frac{-20+\sqrt{86.4}}{-9.8}
$