0=-x^2+3x+40

Solution for 0=-x^2+3x+40 equation:

0=-x^2+3x+40

We move all terms to the left:

0-(-x^2+3x+40)=0

We add all the numbers together, and all the variables

-(-x^2+3x+40)=0

We get rid of parentheses

x^2-3x-40=0

a = 1; b = -3; c = -40;
Δ = b2-4ac
Δ = -32-4·1·(-40)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-13}{2*1}=\frac{-10}{2}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+13}{2*1}=\frac{16}{2}
=8
$