4x^2-9x+12=(x^2+36)

Solution for 4x^2-9x+12=(x^2+36) equation:

4x^2-9x+12=(x^2+36)

We move all terms to the left:

4x^2-9x+12-((x^2+36))=0


We calculate terms in parentheses: -((x^2+36)), so:

(x^2+36)

We get rid of parentheses

x^2+36

Back to the equation:

-(x^2+36)


We get rid of parentheses

4x^2-x^2-9x-36+12=0

We add all the numbers together, and all the variables

3x^2-9x-24=0

a = 3; b = -9; c = -24;
Δ = b2-4ac
Δ = -92-4·3·(-24)
Δ = 369
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:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3\sqrt{41}}{2*3}=\frac{9-3\sqrt{41}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3\sqrt{41}}{2*3}=\frac{9+3\sqrt{41}}{6}
$