2+0.1x-0.15x^2=0

Solution for 2+0.1x-0.15x^2=0 equation:

2+0.1x-0.15x^2=0

a = -0.15; b = 0.1; c = +2;
Δ = b2-4ac
Δ = 0.12-4·(-0.15)·2
Δ = 1.21
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{-(0.1)-\sqrt{1.21}}{2*-0.15}=\frac{-0.1-\sqrt{1.21}}{-0.3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.1)+\sqrt{1.21}}{2*-0.15}=\frac{-0.1+\sqrt{1.21}}{-0.3}
$