4x^2+36x=1x^2+9

Solution for 4x^2+36x=1x^2+9 equation:

4x^2+36x=1x^2+9

We move all terms to the left:

4x^2+36x-(1x^2+9)=0

We get rid of parentheses

4x^2-1x^2+36x-9=0

We add all the numbers together, and all the variables

3x^2+36x-9=0

a = 3; b = 36; c = -9;
Δ = b2-4ac
Δ = 362-4·3·(-9)
Δ = 1404
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:

$\sqrt{\Delta}=\sqrt{1404}=\sqrt{36*39}=\sqrt{36}*\sqrt{39}=6\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-6\sqrt{39}}{2*3}=\frac{-36-6\sqrt{39}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+6\sqrt{39}}{2*3}=\frac{-36+6\sqrt{39}}{6}
$