(5x+4)(4x+25)=139

Solution for (5x+4)(4x+25)=139 equation:

(5x+4)(4x+25)=139

We move all terms to the left:

(5x+4)(4x+25)-(139)=0

We multiply parentheses ..

(+20x^2+125x+16x+100)-139=0

We get rid of parentheses

20x^2+125x+16x+100-139=0

We add all the numbers together, and all the variables

20x^2+141x-39=0

a = 20; b = 141; c = -39;
Δ = b2-4ac
Δ = 1412-4·20·(-39)
Δ = 23001
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(141)-\sqrt{23001}}{2*20}=\frac{-141-\sqrt{23001}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(141)+\sqrt{23001}}{2*20}=\frac{-141+\sqrt{23001}}{40}
$