-4.9x^2+10x+1.6=0

Solution for -4.9x^2+10x+1.6=0 equation:

-4.9x^2+10x+1.6=0

a = -4.9; b = 10; c = +1.6;
Δ = b2-4ac
Δ = 102-4·(-4.9)·1.6
Δ = 131.36
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{-(10)-\sqrt{131.36}}{2*-4.9}=\frac{-10-\sqrt{131.36}}{-9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+\sqrt{131.36}}{2*-4.9}=\frac{-10+\sqrt{131.36}}{-9.8}
$