(4x^2+5x+1)=0

Solution for (4x^2+5x+1)=0 equation:

(4x^2+5x+1)=0

We get rid of parentheses

4x^2+5x+1=0

a = 4; b = 5; c = +1;
Δ = b2-4ac
Δ = 52-4·4·1
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3}{2*4}=\frac{-8}{8}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3}{2*4}=\frac{-2}{8}
=-1/4
$