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

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

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

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

$\sqrt{\Delta}=\sqrt{4}=2$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2}{2*-3}=\frac{-12}{-6}
=+2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2}{2*-3}=\frac{-8}{-6}
=1+1/3
$