3x^2-5=x^2+4x

Solution for 3x^2-5=x^2+4x equation:

3x^2-5=x^2+4x

We move all terms to the left:

3x^2-5-(x^2+4x)=0

We get rid of parentheses

3x^2-x^2-4x-5=0

We add all the numbers together, and all the variables

2x^2-4x-5=0

a = 2; b = -4; c = -5;
Δ = b2-4ac
Δ = -42-4·2·(-5)
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{14}}{2*2}=\frac{4-2\sqrt{14}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{14}}{2*2}=\frac{4+2\sqrt{14}}{4}
$