3x^2+x-1=X^2+4

Solution for 3x^2+x-1=X^2+4 equation:

3x^2+x-1=x^2+4

We move all terms to the left:

3x^2+x-1-(x^2+4)=0

We get rid of parentheses

3x^2-x^2+x-4-1=0

We add all the numbers together, and all the variables

2x^2+x-5=0

a = 2; b = 1; c = -5;
Δ = b2-4ac
Δ = 12-4·2·(-5)
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{41}}{2*2}=\frac{-1-\sqrt{41}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{41}}{2*2}=\frac{-1+\sqrt{41}}{4}
$