-0.02x^2+1.4x+5.3=0

Solution for -0.02x^2+1.4x+5.3=0 equation:

-0.02x^2+1.4x+5.3=0

a = -0.02; b = 1.4; c = +5.3;
Δ = b2-4ac
Δ = 1.42-4·(-0.02)·5.3
Δ = 2.384
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{-(1.4)-\sqrt{2.384}}{2*-0.02}=\frac{-1.4-\sqrt{2.384}}{-0.04}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.4)+\sqrt{2.384}}{2*-0.02}=\frac{-1.4+\sqrt{2.384}}{-0.04}
$