p^2+10p-43=4

Solution for p^2+10p-43=4 equation:

p^2+10p-43=4

We move all terms to the left:

p^2+10p-43-(4)=0

We add all the numbers together, and all the variables

p^2+10p-47=0

a = 1; b = 10; c = -47;
Δ = b2-4ac
Δ = 102-4·1·(-47)
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-12\sqrt{2}}{2*1}=\frac{-10-12\sqrt{2}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+12\sqrt{2}}{2*1}=\frac{-10+12\sqrt{2}}{2}
$