15=3x^2+4x

Solution for 15=3x^2+4x equation:

15=3x^2+4x

We move all terms to the left:

15-(3x^2+4x)=0

We get rid of parentheses

-3x^2-4x+15=0

a = -3; b = -4; c = +15;
Δ = b2-4ac
Δ = -42-4·(-3)·15
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-14}{2*-3}=\frac{-10}{-6}
=1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+14}{2*-3}=\frac{18}{-6}
=-3
$