(p+3)(2p-5)=26

Solution for (p+3)(2p-5)=26 equation:

(p+3)(2p-5)=26

We move all terms to the left:

(p+3)(2p-5)-(26)=0

We multiply parentheses ..

(+2p^2-5p+6p-15)-26=0

We get rid of parentheses

2p^2-5p+6p-15-26=0

We add all the numbers together, and all the variables

2p^2+p-41=0

a = 2; b = 1; c = -41;
Δ = b2-4ac
Δ = 12-4·2·(-41)
Δ = 329
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{329}}{2*2}=\frac{-1-\sqrt{329}}{4}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{329}}{2*2}=\frac{-1+\sqrt{329}}{4}
$