-2p^2+7p-6=0

Solution for -2p^2+7p-6=0 equation:

-2p^2+7p-6=0

a = -2; b = 7; c = -6;
Δ = b2-4ac
Δ = 72-4·(-2)·(-6)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*-2}=\frac{-8}{-4}
=+2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*-2}=\frac{-6}{-4}
=1+1/2
$