-4p^2-28p+3=0

Solution for -4p^2-28p+3=0 equation:

-4p^2-28p+3=0

a = -4; b = -28; c = +3;
Δ = b2-4ac
Δ = -282-4·(-4)·3
Δ = 832
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:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-8\sqrt{13}}{2*-4}=\frac{28-8\sqrt{13}}{-8}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+8\sqrt{13}}{2*-4}=\frac{28+8\sqrt{13}}{-8}
$