0=6a^2+53a+40

Solution for 0=6a^2+53a+40 equation:

0=6a^2+53a+40

We move all terms to the left:

0-(6a^2+53a+40)=0

We add all the numbers together, and all the variables

-(6a^2+53a+40)=0

We get rid of parentheses

-6a^2-53a-40=0

a = -6; b = -53; c = -40;
Δ = b2-4ac
Δ = -532-4·(-6)·(-40)
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-53)-43}{2*-6}=\frac{10}{-12}
=-5/6
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-53)+43}{2*-6}=\frac{96}{-12}
=-8
$