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

Solution for (x+4)(4x+1)=0* equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4x^2+x+16x+4)=0

We get rid of parentheses

4x^2+x+16x+4=0

We add all the numbers together, and all the variables

4x^2+17x+4=0

a = 4; b = 17; c = +4;
Δ = b2-4ac
Δ = 172-4·4·4
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-15}{2*4}=\frac{-32}{8}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+15}{2*4}=\frac{-2}{8}
=-1/4
$