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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-4x-5=0

a = 1; b = -4; c = -5;
Δ = b2-4ac
Δ = -42-4·1·(-5)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6}{2*1}=\frac{-2}{2}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6}{2*1}=\frac{10}{2}
=5
$