3x^2+x=1000

Solution for 3x^2+x=1000 equation:

3x^2+x=1000

We move all terms to the left:

3x^2+x-(1000)=0

a = 3; b = 1; c = -1000;
Δ = b2-4ac
Δ = 12-4·3·(-1000)
Δ = 12001
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{12001}}{2*3}=\frac{-1-\sqrt{12001}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{12001}}{2*3}=\frac{-1+\sqrt{12001}}{6}
$