-10p^2+2p+1400=0

Solution for -10p^2+2p+1400=0 equation:

-10p^2+2p+1400=0

a = -10; b = 2; c = +1400;
Δ = b2-4ac
Δ = 22-4·(-10)·1400
Δ = 56004
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:

$\sqrt{\Delta}=\sqrt{56004}=\sqrt{4*14001}=\sqrt{4}*\sqrt{14001}=2\sqrt{14001}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{14001}}{2*-10}=\frac{-2-2\sqrt{14001}}{-20}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{14001}}{2*-10}=\frac{-2+2\sqrt{14001}}{-20}
$