(0.3+x)-0.140(0.5-x)(0.5-x)=0

Solution for (0.3+x)-0.140(0.5-x)(0.5-x)=0 equation:

(0.3+x)-0.140(0.5-x)(0.5-x)=0

We add all the numbers together, and all the variables

(x+0.3)-0.140(-1x+0.5)(-1x+0.5)=0

We get rid of parentheses

x-0.140(-1x+0.5)(-1x+0.5)+0.3=0

We multiply parentheses ..

-0.140(+x^2-0.5x-0.5x+0.25)+x+0.3=0

We multiply parentheses

-0.14x^2-0x-0x+x-0.035+0.3=0

We add all the numbers together, and all the variables

-0.14x^2-1x+0.265=0

a = -0.14; b = -1; c = +0.265;
Δ = b2-4ac
Δ = -12-4·(-0.14)·0.265
Δ = 1.1484
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)-\sqrt{1.1484}}{2*-0.14}=\frac{1-\sqrt{1.1484}}{-0.28}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{1.1484}}{2*-0.14}=\frac{1+\sqrt{1.1484}}{-0.28}
$