0=(x^2)+(0.32x)-0.84

Solution for 0=(x^2)+(0.32x)-0.84 equation:

0=(x^2)+(0.32x)-0.84

We move all terms to the left:

0-((x^2)+(0.32x)-0.84)=0

We add all the numbers together, and all the variables

-(x^2+(+0.32x)-0.84)+0=0

We add all the numbers together, and all the variables

-(x^2+(+0.32x)-0.84)=0


We calculate terms in parentheses: -(x^2+(+0.32x)-0.84), so:

x^2+(+0.32x)-0.84

We get rid of parentheses

x^2+0.32x-0.84

Back to the equation:

-(x^2+0.32x-0.84)


We get rid of parentheses

-x^2-0.32x+0.84=0

We add all the numbers together, and all the variables

-1x^2-0.32x+0.84=0

a = -1; b = -0.32; c = +0.84;
Δ = b2-4ac
Δ = -0.322-4·(-1)·0.84
Δ = 3.4624
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{-(-0.32)-\sqrt{3.4624}}{2*-1}=\frac{0.32-\sqrt{3.4624}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.32)+\sqrt{3.4624}}{2*-1}=\frac{0.32+\sqrt{3.4624}}{-2}
$