.8636x^2+.2073x+.0124=x^2

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Solution for .8636x^2+.2073x+.0124=x^2 equation: 



.8636x^2+.2073x+.0124=x^2

We move all terms to the left:

.8636x^2+.2073x+.0124-(x^2)=0

determiningTheFunctionDomain
.8636x^2-x^2+.2073x+.0124=0

We add all the numbers together, and all the variables

-0.1364x^2+.2073x+0.0124=0

a = -0.1364; b = .2073; c = +0.0124;
Δ = b2-4ac
Δ = .20732-4·(-0.1364)·0.0124
Δ = 0.04973873
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{-(.2073)-\sqrt{0.04973873}}{2*-0.1364}=\frac{-0.2073-\sqrt{0.04973873}}{-0.2728}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(.2073)+\sqrt{0.04973873}}{2*-0.1364}=\frac{-0.2073+\sqrt{0.04973873}}{-0.2728}
$

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