3x2+13x=10

Solution for 3x2+13x=10 equation:

3x^2+13x=10

We move all terms to the left:

3x^2+13x-(10)=0

a = 3; b = 13; c = -10;
Δ = b2-4ac
Δ = 132-4·3·(-10)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-17}{2*3}=\frac{-30}{6}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*3}=\frac{4}{6}
=2/3
$