-2(b^2+3b-2)=0

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

-2(b^2+3b-2)=0

We multiply parentheses

-2b^2-6b+4=0

a = -2; b = -6; c = +4;
Δ = b2-4ac
Δ = -62-4·(-2)·4
Δ = 68
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{17}}{2*-2}=\frac{6-2\sqrt{17}}{-4}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{17}}{2*-2}=\frac{6+2\sqrt{17}}{-4}
$