5x^2+36x+28=3x^2

Solution for 5x^2+36x+28=3x^2 equation:

5x^2+36x+28=3x^2

We move all terms to the left:

5x^2+36x+28-(3x^2)=0

determiningTheFunctionDomain
5x^2-3x^2+36x+28=0

We add all the numbers together, and all the variables

2x^2+36x+28=0

a = 2; b = 36; c = +28;
Δ = b2-4ac
Δ = 362-4·2·28
Δ = 1072
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:

$\sqrt{\Delta}=\sqrt{1072}=\sqrt{16*67}=\sqrt{16}*\sqrt{67}=4\sqrt{67}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{67}}{2*2}=\frac{-36-4\sqrt{67}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{67}}{2*2}=\frac{-36+4\sqrt{67}}{4}
$