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

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

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

We move all terms to the left:

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

We multiply parentheses

-(-3x^2+2x+5)+3x^2+15x=0

We get rid of parentheses

3x^2+3x^2-2x+15x-5=0

We add all the numbers together, and all the variables

6x^2+13x-5=0

a = 6; b = 13; c = -5;
Δ = b2-4ac
Δ = 132-4·6·(-5)
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*6}=\frac{-30}{12}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*6}=\frac{4}{12}
=1/3
$