0=5x^2+21x+4

Solution for 0=5x^2+21x+4 equation:

0=5x^2+21x+4

We move all terms to the left:

0-(5x^2+21x+4)=0

We add all the numbers together, and all the variables

-(5x^2+21x+4)=0

We get rid of parentheses

-5x^2-21x-4=0

a = -5; b = -21; c = -4;
Δ = b2-4ac
Δ = -212-4·(-5)·(-4)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-19}{2*-5}=\frac{2}{-10}
=-1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+19}{2*-5}=\frac{40}{-10}
=-4
$