40+2x-2x^2=0

Solution for 40+2x-2x^2=0 equation:

40+2x-2x^2=0

a = -2; b = 2; c = +40;
Δ = b2-4ac
Δ = 22-4·(-2)·40
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-18}{2*-2}=\frac{-20}{-4}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+18}{2*-2}=\frac{16}{-4}
=-4
$