(4x+1)*(x-5)=25

Solution for (4x+1)*(x-5)=25 equation:

(4x+1)(x-5)=25

We move all terms to the left:

(4x+1)(x-5)-(25)=0

We multiply parentheses ..

(+4x^2-20x+x-5)-25=0

We get rid of parentheses

4x^2-20x+x-5-25=0

We add all the numbers together, and all the variables

4x^2-19x-30=0

a = 4; b = -19; c = -30;
Δ = b2-4ac
Δ = -192-4·4·(-30)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-29}{2*4}=\frac{-10}{8}
=-1+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+29}{2*4}=\frac{48}{8}
=6
$