-0.02x^2+3.4x-16=-0

Solution for -0.02x^2+3.4x-16=-0 equation:

-0.02x^2+3.4x-16=-0

a = -0.02; b = 3.4; c = -16;
Δ = b2-4ac
Δ = 3.42-4·(-0.02)·(-16)
Δ = 10.28
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{-(3.4)-\sqrt{10.28}}{2*-0.02}=\frac{-3.4-\sqrt{10.28}}{-0.04}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.4)+\sqrt{10.28}}{2*-0.02}=\frac{-3.4+\sqrt{10.28}}{-0.04}
$