4p^2-20p+100=77

Solution for 4p^2-20p+100=77 equation:

4p^2-20p+100=77

We move all terms to the left:

4p^2-20p+100-(77)=0

We add all the numbers together, and all the variables

4p^2-20p+23=0

a = 4; b = -20; c = +23;
Δ = b2-4ac
Δ = -202-4·4·23
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{2}}{2*4}=\frac{20-4\sqrt{2}}{8}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{2}}{2*4}=\frac{20+4\sqrt{2}}{8}
$