(7x-2)(0.3+x)=0

Solution for (7x-2)(0.3+x)=0 equation:

(7x-2)(0.3+x)=0

We add all the numbers together, and all the variables

(7x-2)(x+0.3)=0

We multiply parentheses ..

(+7x^2+2.1x-2x-0.6)=0

We get rid of parentheses

7x^2+2.1x-2x-0.6=0

We add all the numbers together, and all the variables

7x^2+0.1x-0.6=0

a = 7; b = 0.1; c = -0.6;
Δ = b2-4ac
Δ = 0.12-4·7·(-0.6)
Δ = 16.81
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{16.81}}{2*7}=\frac{-0.1-\sqrt{16.81}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.1)+\sqrt{16.81}}{2*7}=\frac{-0.1+\sqrt{16.81}}{14}
$