3x^2+1=x^2-5x

Solution for 3x^2+1=x^2-5x equation:

3x^2+1=x^2-5x

We move all terms to the left:

3x^2+1-(x^2-5x)=0

We get rid of parentheses

3x^2-x^2+5x+1=0

We add all the numbers together, and all the variables

2x^2+5x+1=0

a = 2; b = 5; c = +1;
Δ = b2-4ac
Δ = 52-4·2·1
Δ = 17
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{-(5)-\sqrt{17}}{2*2}=\frac{-5-\sqrt{17}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{17}}{2*2}=\frac{-5+\sqrt{17}}{4}
$