x^2-3x+(9/4)=6

Solution for x^2-3x+(9/4)=6 equation:

x^2-3x+(9/4)=6

We move all terms to the left:

x^2-3x+(9/4)-(6)=0

determiningTheFunctionDomain
x^2-3x-6+(9/4)=0

We add all the numbers together, and all the variables

x^2-3x-6+(+9/4)=0

We get rid of parentheses

x^2-3x-6+9/4=0

We multiply all the terms by the denominator


x^2*4-3x*4+9-6*4=0

We add all the numbers together, and all the variables

x^2*4-3x*4-15=0

Wy multiply elements

4x^2-12x-15=0

a = 4; b = -12; c = -15;
Δ = b2-4ac
Δ = -122-4·4·(-15)
Δ = 384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8\sqrt{6}}{2*4}=\frac{12-8\sqrt{6}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8\sqrt{6}}{2*4}=\frac{12+8\sqrt{6}}{8}
$