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

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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 $

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