(3p^2+3p-2)+-6p^2+5p-5=0

Solution for (3p^2+3p-2)+-6p^2+5p-5=0 equation:

(3p^2+3p-2)+-6p^2+5p-5=0

We add all the numbers together, and all the variables

-6p^2+5p+(3p^2+3p-2)=0

We get rid of parentheses

-6p^2+3p^2+5p+3p-2=0

We add all the numbers together, and all the variables

-3p^2+8p-2=0

a = -3; b = 8; c = -2;
Δ = b2-4ac
Δ = 82-4·(-3)·(-2)
Δ = 40
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{10}}{2*-3}=\frac{-8-2\sqrt{10}}{-6}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{10}}{2*-3}=\frac{-8+2\sqrt{10}}{-6}
$