(5x+1)(x+4)=

Solution for (5x+1)(x+4)= equation:

(5x+1)(x+4)=

We move all terms to the left:

(5x+1)(x+4)-()=0

We add all the numbers together, and all the variables

(5x+1)(x+4)=0

We multiply parentheses ..

(+5x^2+20x+x+4)=0

We get rid of parentheses

5x^2+20x+x+4=0

We add all the numbers together, and all the variables

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