3x^2+x-5=1

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

3x^2+x-5=1

We move all terms to the left:

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

We add all the numbers together, and all the variables

3x^2+x-6=0

a = 3; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·3·(-6)
Δ = 73
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{73}}{2*3}=\frac{-1-\sqrt{73}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{73}}{2*3}=\frac{-1+\sqrt{73}}{6}
$