4b^2-19b+17=0

Solution for 4b^2-19b+17=0 equation:

4b^2-19b+17=0

a = 4; b = -19; c = +17;
Δ = b2-4ac
Δ = -192-4·4·17
Δ = 89
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{89}}{2*4}=\frac{19-\sqrt{89}}{8}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{89}}{2*4}=\frac{19+\sqrt{89}}{8}
$