-2b^2+4b+48=0

Solution for -2b^2+4b+48=0 equation:

-2b^2+4b+48=0

a = -2; b = 4; c = +48;
Δ = b2-4ac
Δ = 42-4·(-2)·48
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20}{2*-2}=\frac{-24}{-4}
=+6
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*-2}=\frac{16}{-4}
=-4
$