-2b^2-3b+5=0

Solution for -2b^2-3b+5=0 equation:

-2b^2-3b+5=0

a = -2; b = -3; c = +5;
Δ = b2-4ac
Δ = -32-4·(-2)·5
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*-2}=\frac{-4}{-4}
=1
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*-2}=\frac{10}{-4}
=-2+1/2
$