-9b^2+26b-16=0

Solution for -9b^2+26b-16=0 equation:

-9b^2+26b-16=0

a = -9; b = 26; c = -16;
Δ = b2-4ac
Δ = 262-4·(-9)·(-16)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-10}{2*-9}=\frac{-36}{-18}
=+2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+10}{2*-9}=\frac{-16}{-18}
=8/9
$