L(p)=3p2+25-140

Solution for L(p)=3p2+25-140 equation:

(L)=3L^2+25-140

We move all terms to the left:

(L)-(3L^2+25-140)=0

We get rid of parentheses

-3L^2+L-25+140=0

We add all the numbers together, and all the variables

-3L^2+L+115=0

a = -3; b = 1; c = +115;
Δ = b2-4ac
Δ = 12-4·(-3)·115
Δ = 1381
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$L_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$L_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$L_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{1381}}{2*-3}=\frac{-1-\sqrt{1381}}{-6}
$
$L_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1381}}{2*-3}=\frac{-1+\sqrt{1381}}{-6}
$