L=p^2+10p-21

Solution for L=p^2+10p-21 equation:

=L^2+10L-21

We move all terms to the left:

-(L^2+10L-21)=0

We get rid of parentheses

-L^2-10L+21=0

We add all the numbers together, and all the variables

-1L^2-10L+21=0

a = -1; b = -10; c = +21;
Δ = b2-4ac
Δ = -102-4·(-1)·21
Δ = 184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$L_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$L_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$L_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{46}}{2*-1}=\frac{10-2\sqrt{46}}{-2}
$
$L_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{46}}{2*-1}=\frac{10+2\sqrt{46}}{-2}
$