0=-3x2+12x+40

Solution for 0=-3x2+12x+40 equation:

0=-3x^2+12x+40

We move all terms to the left:

0-(-3x^2+12x+40)=0

We add all the numbers together, and all the variables

-(-3x^2+12x+40)=0

We get rid of parentheses

3x^2-12x-40=0

a = 3; b = -12; c = -40;
Δ = b2-4ac
Δ = -122-4·3·(-40)
Δ = 624
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{624}=\sqrt{16*39}=\sqrt{16}*\sqrt{39}=4\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{39}}{2*3}=\frac{12-4\sqrt{39}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{39}}{2*3}=\frac{12+4\sqrt{39}}{6}
$