0=5x^2+40x-27

Solution for 0=5x^2+40x-27 equation:

0=5x^2+40x-27

We move all terms to the left:

0-(5x^2+40x-27)=0

We add all the numbers together, and all the variables

-(5x^2+40x-27)=0

We get rid of parentheses

-5x^2-40x+27=0

a = -5; b = -40; c = +27;
Δ = b2-4ac
Δ = -402-4·(-5)·27
Δ = 2140
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:

$\sqrt{\Delta}=\sqrt{2140}=\sqrt{4*535}=\sqrt{4}*\sqrt{535}=2\sqrt{535}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{535}}{2*-5}=\frac{40-2\sqrt{535}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{535}}{2*-5}=\frac{40+2\sqrt{535}}{-10}
$