-8p^2+10p-3=0

Solution for -8p^2+10p-3=0 equation:

-8p^2+10p-3=0

a = -8; b = 10; c = -3;
Δ = b2-4ac
Δ = 102-4·(-8)·(-3)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2}{2*-8}=\frac{-12}{-16}
=3/4
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2}{2*-8}=\frac{-8}{-16}
=1/2
$