47p^2+43p=0

Solution for 47p^2+43p=0 equation:

47p^2+43p=0

a = 47; b = 43; c = 0;
Δ = b2-4ac
Δ = 432-4·47·0
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-43}{2*47}=\frac{-86}{94}
=-43/47
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+43}{2*47}=\frac{0}{94}
=0
$