2x^2-5=-x^2+4x-1

Solution for 2x^2-5=-x^2+4x-1 equation:

2x^2-5=-x^2+4x-1

We move all terms to the left:

2x^2-5-(-x^2+4x-1)=0

We get rid of parentheses

2x^2+x^2-4x+1-5=0

We add all the numbers together, and all the variables

3x^2-4x-4=0

a = 3; b = -4; c = -4;
Δ = b2-4ac
Δ = -42-4·3·(-4)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*3}=\frac{-4}{6}
=-2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*3}=\frac{12}{6}
=2
$