3x2+x=10

Solution for 3x2+x=10 equation:

3x^2+x=10

We move all terms to the left:

3x^2+x-(10)=0

a = 3; b = 1; c = -10;
Δ = b2-4ac
Δ = 12-4·3·(-10)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-11}{2*3}=\frac{-12}{6}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*3}=\frac{10}{6}
=1+2/3
$