(-4p^2-p+9)+(p^2+3p-1)=0

Solution for (-4p^2-p+9)+(p^2+3p-1)=0 equation:

(-4p^2-p+9)+(p^2+3p-1)=0

We get rid of parentheses

-4p^2+p^2-p+3p+9-1=0

We add all the numbers together, and all the variables

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

a = -3; b = 2; c = +8;
Δ = b2-4ac
Δ = 22-4·(-3)·8
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*-3}=\frac{-12}{-6}
=+2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*-3}=\frac{8}{-6}
=-1+1/3
$