-w^2+4w-3=0

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

-w^2+4w-3=0

We add all the numbers together, and all the variables

-1w^2+4w-3=0

a = -1; b = 4; c = -3;
Δ = b2-4ac
Δ = 42-4·(-1)·(-3)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2}{2*-1}=\frac{-6}{-2}
=+3
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2}{2*-1}=\frac{-2}{-2}
=1
$